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NC 8th Grade Chemistry

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Bewick Edge Forsythe Parsons

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Sharon Bewick, Jonathan Edge, Therese Forsythe, Richard Parsons

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Contents

1 The Science of Chemistry 1

1.1 The Scientific Method 1

1.2 Chemistry in History 29

1.3 Chemistry is a Science of Materials 35

1.4 Matter 42

1.5 Energy 48

2 Chemistry - A Physical Science 57

2.1 Measurements in Chemistry 57

2.2 Using Measurements 62

2.3 Using Mathematics in Chemistry 66

2.4 Using Algebra in Chemistry 69

3 Chemistry in the Laboratory 75

3.1 Making Observations 75

3.2 Making Measurements 78

3.3 Using Data 89

3.4 How Scientists Use Data 110

4 Atomic Theory 115

4.1 Early Development of a Theory 115

4.2 Further Understanding of the Atom 123

4.3 Atomic Terminology 133

5 The Bohr Model 148

5.1 The Wave Form of Light 148

5.2 The Dual Nature of Light 154

5.3 Light and the Atomic Spectra 164

5.4 The Bohr Model 171

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6 Quantum Mechanics Model of

the Atom 182

6.1 The Wave Particle Duality 182

6.2 Schrodinger's Wave Functions 191

6.3 Heisenberg's Contribution 198

6.4 Quantum Numbers 203

6.5 Shapes of Atomic Orbitals 213

7 Electron Configurations for

Atoms 224

7.1 The Electron Spin Quantum Number 224

7.2 Pauli Exclusion Principle 233

7.3 Aufbau Principle 236

7.4 Writing Electron Configurations 240

8 Electron Configurations and

the Periodic Table 246

8.1 Electron Configurations of Main Group Elements 246

8.2 Orbital Configurations 257

8.3 The Periodic Table and Electron Configurations 263

9 Relationships Between the

Elements 272

9.1 Families on Periodic Table 272

9.2 Electron Configurations 281

9.3 Lewis Electron Dot Diagrams 291

9.4 Chemical Family Members Have Similar Properties 297

9.5 Transition Elements 300

9.6 Lanthanides and Actinides 305

10 Trends on the Periodic Table 310

10.1 Atomic Size 310

10.2 Ionization Energy 321

10.3 Electron Affinity 327

11 Covalent Bonding 331

11.1 The Covalent Bond 331

11.2 Atoms that Form Covalent Bonds 335

11.3 Naming Covalent Compounds 342

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12 Reactions 345

12.1 Chemical Equations 345

12.2 Balancing Equations 349

12.3 Types of Reactions 356

13 The Liquid State 372

13.1 The Properties of Liquids 372

13.2 Forces of Attraction 374

13.3 Vapor Pressure 382

13.4 Boiling Point 389

13.5 Heat of Vaporization 392

14 The Solid State 396

14.1 The Molecular Arrangement in Solids Controls Solid Characteristics 396

14.2 Melting 397

14.3 Types of Forces of Attraction for Solids 402

14.4 Phase Diagrams 410

15 The Solution Process 415

15.1 What Are Solutions? 415

15.2 Why Solutions Occur 418

15.3 Solution Terminology 421

15.4 Measuring Concentration 424

15.5 Solubility Graphs 429

15.6 Factors Affecting Solubility 435

15.7 Colligative Properties 439

15.8 Colloids 445

15.9 Separating Mixtures 447

16 Ions in Solution 452

16.1 Ionic Solutions 452

16.2 Covalent Compounds in Solution 458

16.3 Reactions Between Ions in Solutions 462

17 Acids and Bases 481

17.1 Arrhenius Acids 481

17.2 Strong and Weak Acids 486

17.3 Arrhenius Bases 489

17.4 Salts 493

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17.5 pH 502

17.6 Weak Acid/Base Equilibria 510

17.7 Br0nsted Lowry Acids-Bases 516

17.8 Lewis Acids and Bases 522

18 Water, pH and Titration 527

18.1 Water Ionizes 527

18.2 Indicators 531

18.3 Titrations 541

18.4 Buffers 559

19 Radioactivity and the Nucleus 564

19.1 Discovery of Radioactivity 564

19.2 Nuclear Notation 567

19.3 Nuclear Force 569

19.4 Nuclear Disintegration 572

19.5 Nuclear Equations 576

19.6 Radiation Around Us 581

19.7 Applications of Nuclear Energy 585

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Chapter 1

The Science of Chemistry

1.1 The Scientific Method

Lesson Objectives

• Describe the steps involved in the scientific method.

• Appreciate the value of the scientific method.

• Recognize that in some cases not all the steps in the scientific method occur, or they do not occur in
any specific order.

• Explain the necessity for experimental controls.

• Recognize the components in an experiment that represent experimental controls.

Introduction

"What hopes and fears does this scientific method imply for mankind? I do not think that this is the right
way to put the question. Whatever this tool in the hand of man will produce depends entirely on the
nature of the goals alive in this mankind. Once the goals exist, scientific method furnishes means to realize
them. Yet it cannot furnish the very goals. The scientific method itself would not have led anywhere, it
would not even have been born without a passionate striving for clear understanding." - Albert Einstein

Historical Comparisons
Introduction to Science

What is science? Is it a list of marvelous inventions and how they work? Or is it a list of theories about
matter and energy and biological systems? Or is science a subject that you learn by carrying out activities
in a laboratory? Science is all of these, but it is also something even more basic. Science is a method of
thinking that allows us to discover how the world around us works.

To begin this study of one form of science, we will review the last 3,000 years in the history of human
transportation, communication, and medicine. The following summary lists humankind's accomplishments
in these areas during three periods in the last 3,000 years.

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Transportation in 1000 B.C.

In 1000 B.C., people could transport themselves and their goods by walking, riding an animal, or by riding
in a cart pulled by an animal (Figure ??). Crossing water, people could paddle a boat or have an animal
walk beside the river and pull the boat (Figure 1.1). These methods of transportation required muscle
power, either human muscles or animal muscles.

Figure 1.1: A photo of a wooden model of a Greek ship that has both sails and oars.

A few societies had designed rowboats or sailboats, which used muscle power or the force of the wind to
move the boat. These early means of transportation were very limited in terms of speed and therefore,
also limited in terms of distances traveled. The sail and rowboats were used on rivers and inland seas, but
were not ocean-going vessels.

Transportation in 1830

By the year 1830, people were still walking and riding in carts pulled by animals. Iron ore was moved
along canals by animals pulling barges. American pioneers crossed the United States in covered wagons
pulled by animals (Figure 1.2). Large cities had streetcars pulled by horses (Figure 1.3). Ocean crossing
was accomplished in sailing ships. The only improvement in transportation was the addition of springs

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and padded seats to carts and wagons to make the ride less jolting. In the period from 1000 B.C. to 1830,
a span of 2,830 years (about 100 generations of people), there were no significant changes in the mode
of human transportation.

Figure 1.2: A covered wagon of the type used by pioneers to cross the US in the mid- 1800s.

Figure 1.3: The first horse-drawn street car in Seattle, Washington in 1884.

Transportation in 1995

By the year 1995, steam engines, gasoline engines, automobiles, propeller-driven and jet engines, loco-
motives, nuclear-powered ships, and inter-planetary rocket ships were invented (Figure 1.4). In all in-
dustrialized countries, almost anyone could own an automobile and travel great distances in very short
times.

In the mid- 1800s, several months were required to travel from Missouri to California by covered wagon
and the trip was made at considerable risk to the traveler's life. In 1995, an average family could travel
this same distance easily in two days and in relative safety. An ordinary person in 1995 probably traveled
a greater distance in one year than an ordinary person in 1830 did in an entire lifetime. The significant
changes in the means of transportation in the 165 years between 1830 and 1995 (perhaps 5 generations)
were phenomenal.

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Figure 1.4: A modern jetliner.

Communication in 1000 B.C.

Figure 1.5: Shopping list chiseled on a rock.

Essentially, people's only means of communicating over large distances (more than 15 miles) in 1000 B.C.
was to send hand-carried messages (Figure 1.5). Some societies, for short distances, had developed the use
of smoke signals, light signals, or drum signals, but these methods were useless for long distances. Since
the means of communicating required hand-carried messages, the speed of communication was limited by
the speed of transportation. Sending messages over distances of 1,000 miles could require several weeks
and even then delivery was not guaranteed.

Communication in 1830

By the year 1830, people's means of communication over large distances was still the hand-carried message.
While the paper and ink used to write the message had been improved, it still had to be hand-carried.
In the United States, communication between New York and San Francisco required more than a month.
When a new president was elected, Californians would not know who it was for a couple of months after
the election.

For a short period of time, the Pony Express was set up and could deliver a letter from St. Louis, Missouri
to Sacramento, California in eleven days, which was amazing at the time (Figure 1.6). The means of
communication in 1830 was essentially the same as in 1000 B.C.

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Figure 1.6: A pony express rider, circa 1861.

Communication in 1995

Figure 1.7: A modern cell phone.

By the year 1995, the telegraph, telephone, radio, television, optical fibers, and communication satellites
were invented (Figure 1.7). People could communicate almost anywhere in the industrialized world in-
stantaneously. Now, when a U.S. president is elected, people around the globe know the name of the new
president the instant the last vote is counted. Astronauts communicate directly between the earth and
the moon. An ordinary person in an industrialized country can speak with people around the world while
simultaneously watching events occur in real time globally. There have been truly extraordinary changes
in people's ability to communicate in the last 165 years.

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Medical Treatment in 1000 B.C.

Figure 1.8: Physician letting blood from a patient.

Medical treatment in 1000 B.C. consisted of a few natural herbs and some superstitious chants and dances.
The most advanced societies used both sorcerers and herbalists for medical treatment. Some of the natural
herbs helped the patient and some did not. Cleaning and bandaging wounds decreased opportunity for
infection while some herbs such as sesame oil demonstrated moderate antiseptic properties. Dances, chants,
incense burning, and magic spells were absolutely useless in curing illnesses. At some point in time,
bloodletting was added to the physician's repertoire (Figure 1.8). Bloodletting was accomplished by
cutting the patient and allowing the blood to drip out or by applying leeches (which doctors often carried
with them). However, bloodletting was not helpful to the patient, and in many cases, it was harmful.
Bloodletting was flourishing by 500 B.C. and was carried out by both surgeons and barbers. It wasn't
until around 1875 that bloodletting was established as quackery.

In those times, for an ordinary person, broken bones went unset and injuries like deep cuts or stab wounds
were often fatal due to infection. Infant mortality was high and it was common for at least one child in a
family die before adulthood. The death of the mother in childbirth was also quite common.

In the Middle Ages, knowledge of germs, hygiene, and contagion was non-existent. People who were
seriously ill might have their disease blamed on the planets going out of line (astrology) or "bad odors,"
or retribution for sins, or an imbalance in body fluids. Cures could involve anything from magic spells,
bleeding, sweating, and vomiting to re-balance bodily fluids. Between 1340 AD and 1348 AD, the Black
Death (bubonic plague) was responsible for killing in the vicinity of half the population of Europe. The
bacterium causing the disease was carried by fleas, but, of course, none of this was known by the physicians
of the time. Efforts to stop the plague included burning incense to eliminate "bad odors," causing loud
noises to chase the plague away (the constant ringing of bells or firing of canon), and a number of people
used self-flagellation to attempt to cure the disease.

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Medical Treatment in 1830

Medical treatment in 1830 remained in the form of natural herbs and bloodletting. During this time, the
ability to set broken bones and to amputate limbs was also developed. Amputation saved many lives from
infection and gangrene. Gangrene occurs when the blood supply to tissue is interrupted and the tissue
dies. The dead tissue remains part of the body, invites infection, and causes death as the poisons from the
rotting tissue are carried through the body. Once gangrene afflicted an arm or leg, the poison from the
limb would eventually kill the patient. During the American Civil War (1861 - 1863), a common means
of treatment for wounds in field hospitals was amputation. Along with amputation was the ability to
cauterize wounds to stop bleeding.

Even though bloodletting did not help patients, it continued in use through 1830. There is a tale (which
may or may not be true) that George Washington was suffering from pneumonia and his doctors removed
so much blood trying to cure him that they actually caused his death.

Medical Treatment in 1995

Figure 1.9: Receiving a vaccination.

By 1995, medical science had discovered chemical medicines, antiseptic procedures, surgery, and probably
most important of all, vaccination . . . the ability to prevent disease rather than cure it after it had been
contracted (Figure 1.9).

Diseases that had killed and crippled hundreds of thousands of people in the past are seldom heard of
today (polio, smallpox, cholera, bubonic plague, etc.). These diseases have been controlled by scientific
understanding of their causes and carriers and by vaccination. Average life expectancy has nearly doubled
in the last 165 years. Both infant mortality and death during childbirth rates have dropped to less than
25% of what they were in 1830.

Methods of Learning About Nature
Opinion, Authority, and Superstition

Why did humans make so little progress in the 2,800 years before 1830 and then such incredible progress
in the 160 years after 1830?

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Socrates (469 B.C. - 399 B.C.), Plato (427 B.C. - 347 B.C.), and Aristotle (384 B.C. - 322 B.C.) are among
the most famous of the Greek philosophers. Plato was a student of Socrates and Aristotle was a student of
Plato. These three were probably the greatest thinkers of their time. Aristotle's views on physical science
profoundly shaped medieval scholarship and his influence extended into the Renaissance (14'' 1 century).
Aristotle's opinions were the authority on nature until well into the 1300s.

Unfortunately, many of Aristotle's opinions were wrong. It is not intended here to denigrate Aristotle's
intelligence; he was without doubt a brilliant man. It was simply that he was using a method for determining
the nature of the physical world that is inadequate for that task. The philosopher's method was logical
thinking, not making observations on the natural world. This led to many errors in Aristotle's thinking on
nature. Let's consider just two of Aristotle's opinions as examples.

In Aristotle's opinion, men were bigger and stronger than women, and therefore, it was logical that men
would have more teeth than women. Therefore, Aristotle concluded it was a true fact that men had more
teeth than women. Apparently, it never entered his mind to actually look into the mouths of both genders
and count their teeth. Had he done so, he would have found that men and women have exactly the same
number of teeth.

QS>

Figure Galileo dropping balls off the Leaning Tower of Pisa.

In terms of physical science, Aristotle thought about dropping two balls of exactly the same size and shape
but of different masses to see which one would strike the ground first. In his mind, it was clear that the
heavier ball would fall faster than the lighter one and he concluded that this was a law of nature. Once
again, he did not consider doing an experiment to see which ball fell faster. It was logical to him, and
in fact, it still seems logical. If someone told you that the heavier ball would fall faster, you would have
no reason to disbelieve it. In fact, it is not true and the best way to prove this is to try it. Eighteen
centuries later, Galileo decided to actually get two balls of different masses, but with the same size and
shape, and drop them off a building (legend says the Leaning Tower of Pisa), and actually see which one
hit the ground first. When Galileo actually did the experiment, he discovered, by observation, that the
two balls hit the ground at exactly the same time . . . Aristotle's opinion was, once again, wrong.

In the 16 and 17 r/l centuries, innovative thinkers were developing a new way to discover the nature of the
world around them. They were developing a method that relied upon making observations of phenomena
and insisting that their explanations of the nature of the phenomena corresponded to the observations they
made. In order to do this, they had to overcome the opinions of the ancient Greeks, the authority of the
church, and the superstitions of ordinary people.

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In the opinion of the ancient Greeks, the earth was the center of the universe and did not move, while the
sun, moon, planets, and stars revolved around the Earth in orbits. The astronomer Ptolemy (around 150
A.D.) observed the positions of the planets and recognized that the positions where he observed the planets
did not match up with the positions predicted by the orbits of the Greeks. Ptolemy designed new orbits
that had circles within circles and complicated retrograde motion (planets moving backward in their orbits
at certain times). His descriptions came closer but still could not accurately predict where the heavenly
orbs would be on a given night. It wasn't until Nicolaus Copernicus (1473 - 1543) suggested a heliocentric
(sun-centered) system that the positions of the planets came close to matching predictions. Copernicus was
hesitant to publish his ideas - some say because he feared ridicule from his peers and others say because
he feared persecution by the church - but eventually, he sent his work for publication just before his death.

The publication of Copernicus' heliocentric theory didn't seem to cause much controversy for the next 50
years until the idea was supported by a scientist named Giordano Bruno who was promptly prosecuted
and burned at the stake by Cardinal Bellarmini in 1600. The most famous supporter of the Copernican
system was Galileo Galilei (1564 - 1642) who had developed an improved telescope (1610) and turned it
toward the sky. Galileo published a small work describing what he saw with his telescope and how his
observations supported the Copernican theory. The book was banned by the church in 1616 and Galileo
was instructed not to write about the subject any further. In 1632, Galileo published another work, again
supporting the Copernican theory and was arrested by the church, prosecuted, and punished by house
arrest for the remainder of his life.

But the method of learning by experimenting, observing, and hypothesizing had been launched and many
scientists would not turn back. It should be mentioned that the supporters of the methods of opinion,
authority, and superstition did not give and have not given up today. We still have "scientists" claiming
that unsupported opinions are "facts" and we still have people deciding the "truth" about nature by
voting on it. Nor has superstition died. You may remember from your history classes that the pilgrims
of Massachusetts were still drowning and hanging women accused of being witches as late as 1693. It is
easy to think that the people of those times were not very smart, and nowadays, no one would think that
way. However, you should be aware that a student was suspended from school in Tulsa, Oklahoma in 1999
for "casting spells" and a substitute teacher in Florida was fired in 2008 for "wizardry" after performing a
magic trick for his students.

The Scientific Method

Scientists frequently list the scientific method as a series of steps. Other scientists oppose this listing of
steps because not all steps occur in every case and sometimes the steps are out of order. The scientific
method is listed in a series of steps in Table 1.1 and represented in the flowchart (Figure ??) because it
makes it easier to study. You should remember that not all steps occur in every case nor do they always
occur in order.

Table 1.1: The Steps in the Scientific Method

Step Number Step Description

1 Identify the problem or phenomenon that needs ex-
plaining. This is sometimes referred to as "defining
the problem." This activity helps limit the field of
observations.

2 Gather and organize data on the problem. This

step is also known as "making observations."

3 Suggest a possible solution or explanation. A sug-
gested solution is called a hypothesis.

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Table 1.1: (continued)

Step Number

Step Description

4
5

Test the hypothesis by making new observations.
If the new observations support the hypothesis, you
accept the hypothesis for further testing. If the
new observations do not agree with your hypothe-
sis, you discard the hypothesis, add the new data
to your observations list, and return to step 3.

Identify the Problem

Gather Data

Make a Hypothesis

Test the Hypothesis (Experiment)

Does the New Data Agree?

When the results of several experiments support the hypothesis, you might think that the work is finished.
However, for a hypothesis to be useful, it must withstand repeated testing. Other scientists must be able
to repeat the experiments using the same materials and conditions and get the same results. Scientists
submit reports of research to other scientists, usually by publishing an article in a scientific journal, so the
work can be verified.

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An Example of the Scientific Method

Suppose you are required to maintain a large campfire and you are completely unfamiliar with the property
of objects that makes them combustible (able to burn). The first step in the scientific method is to define
the problem. What property of objects make them combustible?

The next step is to gather data on the problem. So, you begin to collect objects at random and put them
into the fire. You must keep good records of what objects were tried and whether or not they burned.
Table 1.2 shows a list of organized data (observations):

Table 1.2:

Will Burn Won't Burn

tree limbs rocks

chair legs bricks

pencils marbles

baseball bat hubcaps

The list of organized observations helps because now you can collect only the items on the "will burn" list
and not waste the effort of dragging items that won't burn back to the fire. However, you would soon use
up all the items on the "will burn" list and it is necessary to guess what property the "will burn" objects
have that cause them to burn. If you had that answer, you could bring objects that may not be on the
"will burn" list but that have the "will burn" property and keep the fire going.

The third step in the scientific method is to suggest a hypothesis. Your guess about what property the "will
burn" objects have that makes them combustible is a hypothesis. Suppose you notice that all the items on
the "will burn" list are cylindrical in shape and therefore, you hypothesize that "cylindrical objects burn".
The fourth step in the scientific method is to test your hypothesis. To test this hypothesis, you go out
and collect a group of objects that are cylindrical including iron pipes, soda bottles, broom handles, and
tin cans. When these cylindrical objects are placed in the fire and most of them don't burn, you realize
your hypothesis is not supported by these new observations. The new observations are the test, and your
hypothesis has failed the test. When the new observations fail to support your hypothesis, you reject your
original hypothesis, add your new data to (Table 1.3), and make a new hypothesis based on the updated
observations list. In the schematic diagram of the scientific method, a failed test returns the scientist to
step 3, make a new hypothesis.

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Table 1.3:

Will Burn Won't Burn

tree limbs rocks

chair legs bricks

pencils marbles

baseball bat hubcaps

broom handle iron pipes

soda bottles
tin cans

Suppose your new hypothesis is "wooden objects burn." You will find this hypothesis more satisfactory
since all the wooden object you try will burn. Your confidence will grow that you have discovered a "law
of nature." Even with your somewhat successful theory, you might be ignoring a large stack of old car tires,
objects made of fabric or paper, or perhaps containers of petroleum. You can see that even though you
are quite satisfied with your theory because it does the job you want it to do, you actually do not have a
complete statement on the property of objects that make them burn. So it is with science.

You can see from this example that the "solution" does not become what we think of as a "fact," but rather
becomes a tentatively accepted theory which must undergo continuous testing and perhaps
adjustment. No matter how long a tentative explanation has been accepted, it can be discarded at any
time if contradictory observations are found. As long as the theory is consistent with all observations,
scientists will continue to use it. When a theory is contradicted by observations, it is discarded and
replaced. Even though the terms hypothesis, theory, and fact are used somewhat carelessly at times, a
theory will continue to be used while it is useful and will be called into question when contradictory
evidence is found. Theories never become facts.

There is a common generalization about theories, which says that "theories are much easier to disprove
than to prove." The common example given is a hypothesis that "all swans are white." You may observe a
thousand white swans and every observation of a white swan supports your hypothesis, but it only takes
a single observation of a black swan to disprove the hypothesis. To be an acceptable scientific hypothesis,
observations that disprove the hypothesis must be possible. That is, if every conceivable observation
supports the hypothesis, then it is not an acceptable scientific hypothesis. To be a scientific hypothesis, it
must be possible to refute the concept.

Some Basic Terminology

• A hypothesis is a guess that is made early in the process of trying to explain some set of observations.
There are scientists who object to calling a hypothesis a "guess". The primary basis for the objection
is that someone who has studied the subject under consideration would make a much better guess
than someone who was completely ignorant of the field. Perhaps we should say that a hypothesis is
an "educated guess."

• A theory is an explanation that stands up to everyday use in explaining a set of observations. A
theory is not proven and is not a "fact." A scientific theory must be falsifiable in order to be accepted
as a theory.

• A law describes an observable relationship, that is, observations that occur with a predictable rela-
tionship to each other. It is only after experience shows the law to be valid that it is incorporated
into the field of knowledge.

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The Scientific Revolution

The explosion of achievement in the last 160 years was produced by using a new method for learning
about nature. This sudden and massive achievement in understanding nature is called the Scientific
Revolution and was produced using the scientific method.

The British historian, Herbert Butterfield, wrote a book called The Origins of Modern Science. In the
preface to the book, Butterfield wrote:

The Revolution in science overturned the authority of not only the Middle Ages but of the ancient world .
. . it ended not only in the eclipse of scholastic philosophy but in the destruction of Aristotelian physics.
The Scientific Revolution outshines everything since the rise of Christianity and reduces the Renaissance
and Reformation to the rank of mere episodes . . .

The beginning of the 17 th century is known for the drastic changes that occurred in the European approach
to science during that period and is known as the Scientific Revolution. This term refers to a completely
new era of academic thought in which medieval philosophy was abandoned in favor of innovative methods
offered by Galileo and Newton.

Science is best defined as a careful, disciplined, logical search for knowledge about any and all aspects of
the universe, obtained by examination of the best available evidence and always subject to correction and

improvement upon discovery of better evidence. What's left is magic. And it doesn't work. James

Randi.

What is an Experiment

The scientific method requires than observations be made. Sometimes, the phenomenon we wish to observe
does not occur in nature or if it does, it is inconvenient for us to observe. Therefore, it is more successful
for us to cause the phenomenon to occur at a time and place of our choosing. When we arrange for the
phenomenon to occur at our convenience, we can have all our measuring instruments present and handy to
help us make observations, and we can control other variables. Causing a phenomenon to occur when and
where we want it and under the conditions we want is called an experiment. When scientists conduct
experiments, they are usually seeking new information or trying to verify someone else's data. Classroom
experiments often demonstrate and verify information that is already known but new to the student. When
doing an experiment, it is important to set up the experiment so that relationships can be seen clearly.
This requires what are called experimental controls.

Experimental Controls

Suppose a scientist, while walking along the beach on a very cold day following a rainstorm, observed two
pools of water in bowl shaped rocks near each other. One of the pools was partially covered with ice while
the other pool had no ice on it. The unfrozen pool seemed to be formed from seawater splashing up on
the rock from the surf, but the other pool was too high for sea water to splash in, so it was more likely to
have been formed from rainwater.

The scientist wondered why one pool was partially frozen and not the other since both pools were at the
same temperature. By tasting the water (not a good idea), the scientist determined that the unfrozen pool
tasted saltier than the partially frozen one. The scientist thought perhaps salt water had a lower freezing
point that fresh water and she decided to go home and try an experiment to see if this were true. So far,
the scientist has identified a question, gathered a small amount of data, and suggested a hypothesis. In
order to test this hypothesis, the scientist will conduct an experiment during which she can make accurate
observations.

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For the experiment, the scientist prepared two identical containers of fresh water and added some salt to
one of them (Figure ??). A thermometer was placed in each liquid and these were put in a freezer. The
scientist then observed the conditions and temperatures of the two liquids at regular intervals.

The Temperature and Condition of Fresh Water in a Freezer

Time (minu

tes) 5

10 15

Temp., C

25 20

15 10

-5

Condition
Temperature

Liquid Liquid
e and Condition of Salt Water

Liquid Liquid
in a Freezer

Liquid

Frozen

Time (minutes) 5

10 15

Temp., C

25 20

15 10

-5

Condition

Liquid Liquid

Liquid

Frozen

The scientist found support for the hypothesis from this experiment; fresh water freezes at a higher tem-
perature than salt water. Much more support would be needed before the scientist would be confident of
this hypothesis. Perhaps she would ask other scientists to verify the work.

In the scientist's experiment, it was necessary that she freeze the salt water and fresh water under exactly
the same conditions. Why? The scientist was testing whether or not the presence of salt in water would
alter its freezing point. It is known that changing air pressure will alter the freezing point of water. In
order to conclude that the presence of the salt was what caused the change in freezing point, all other
conditions had to be identical. The presence of the salt is called the experimental variable because
it is the only thing allowed to change in the two trials. The fresh water part of the experiment is called
the experimental control. In an experiment, there may be only one variable and the purpose of the
control is to guarantee that there is only one variable. The "control" is identical to the "test" except for
the experimental variable. Unless experiments are controlled, the results are not valid.

Suppose you wish to determine which brand of microwave popcorn leaves the fewest unpopped kernels.
You will need a supply of various brands of microwave popcorn to test and you will need a microwave
oven. If you used different brands of microwave ovens with different brands of popcorn, the percentage
of unpopped kernels could be caused by the different brands of popcorn, but it could also be caused by
the different brands of ovens. Under such circumstances, the experimenter would not be able to conclude
confidently whether the popcorn or the oven caused the difference. To eliminate this problem, you must
use the same microwave oven for every test. By using the same microwave oven, you control the number
of variables in the experiment.

What if you allowed the different samples of popcorn to be cooked at different temperatures? What if
you allowed longer heating periods? In order to reasonably conclude that the change in one variable was

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caused by the change in another specific variable, there must be no other variables in the experiment. All
other variables must be kept constant.

Errors in the Use of the Scientific Method

The scientific method requires the observation of nature and correspondence between the suggested ex-
planation and the observations. That is, the hypothesis must explain all the observations. Therefore,
the scientific method can only work properly when the data (observation list) is not biased. There are
several ways in which a biased set of data can be produced. It is always possible for anyone to make
an error in observation. A balance can be misread or numbers can be transposed when written down.
That is one of the reasons that experiments are run several times and the observations made over and
over again. It is also possible that an unrecognized error is present and produces the same error in every
experiment. For example, a scientist may be attempting to test normal rainwater, but unknown to him
a nearby factory is sending soluble substances out of their smoke stack and the material is contaminating
the scientist's samples. In such a case, the scientist's samples would yield false observations for normal
rainwater. Other scientists reproducing the experiment would collect uncontaminated samples and find
different results. Multiple testing of the experiment would determine which set of data was flawed. Failing
to apply appropriate experimental controls would certainly bias data.

There are also dishonest mistakes that occur when the experimenter collects only supporting data and
excludes contradictory observations. There have even been scientists who faked observations to provide
support for his/her hypothesis. The attractions of fame and fortune can be hard to resist. For all these
reasons, the scientific method requires that experimental results be published and the experiment be
repeated by other scientists.

Lesson Summary

• Before the development of the scientific method, mankind made only slight achievements in the areas
of transportation, communication and medicine.

• Use of the scientific method allowed mankind to make significant achievements in transportation,
communication, and medicine.

• The scientific method has been much more successful than the methods of superstition, opinion, and
authority.

• The steps in the scientific method are:

1. Identify the problem.

2. Gather data (make observations).

3. Suggest a hypothesis.

4. Test the hypothesis (experiment).

5. Continue testing or reject the hypothesis and make a new one.

• Experimental controls are used to make sure that the only variables in an experiment are the ones
being tested.

Review Questions

Use the following paragraph to answer questions 1 and 2. In 1928, Sir Alexander Fleming was studying
Staphylococcus bacteria growing in culture dishes. He noticed that a mold called Penicillium was also
growing in some of the dishes. In Figure ??, Petri dish A represents a dish containing only Staphylococcus

15 www.ckl2.org

bacteria. The red dots in dish B represent Penicillium colonies. Fleming noticed that a clear area existed
around the mold because all the bacteria grown in this area had died. In the culture dishes without the
mold, no clear areas were present. Fleming suggested that the mold was producing a chemical that killed
the bacteria. He decided to isolate this substance and test it to see if it would kill bacteria. Fleming grew
some Penicillium mold in a nutrient broth. After the mold grew in the broth, he removed all the mold
from the broth and added the broth to a culture of bacteria. All the bacteria died.

1. Which of the following statements is a reasonable expression of Fleming's hypothesis?

(a) Nutrient broth kills bacteria.

(b) There are clear areas around the Penicillium mold where Staphylococcus doesn't grow.

(d) Penicillium mold produces a substance that kills Staphylococcus.

(e) Without mold in the culture dish, there were no clear areas in the bacteria.

2. Fleming grew Penicillium in broth, then removed the Penicillium and poured the broth into culture
dishes containing bacteria to see if the broth would kill the bacteria. What step in the scientific
method does this represent?

(a) Collecting and organizing data

(b) Making a hypothesis

(d) Rejecting the old hypothesis and making a new one

(e) None of these

3. A scientific investigation is NOT valid unless every step in the scientific method is present and carried
out in the exact order listed in this chapter.

(a) True

(b) False

4. Which of the following words is closest to the same meaning as hypothesis?

(a) fact

(b) law

(d) suggestion

(e) conclusion

5. Why do scientists sometimes discard theories?

(a) the steps in the scientific method were not followed in order

(b) public opinion disagrees with the theory

(d) contradictory observations are found

(e) congress voted against it

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Gary noticed that two plants which his mother planted on the same day that were the same size when
planted were different in size after three weeks. Since the larger plant was in the full sun all day and the
smaller plant was in the shade of a tree most of the day, Gary believed the sunshine was responsible for
the difference in the plant sizes. In order to test this, Gary bought ten small plants of the size and type.
He made sure they had the same size and type of pot. He also made sure they have the same amount and
type of soil. Then Gary built a frame to hold a canvas roof over five of the plants while the other five
were nearby but out in the sun. Gary was careful to make sure that each plant received exactly the same
amount of water and plant food every day.

6. Which of the following is a reasonable statement of Gary's hypothesis?

(a) Different plants have different characteristics.

(b) Plants that get more sunshine grow larger than plants that get less sunshine.

(d) Plants that don't receive water will die.

(e) Plants that receive the same amount of water and plant food will grow the same amount.

7. What scientific reason might Gary have for insisting that the container size for the all plants be the
same?

(a) Gary wanted to determine if the size of the container would affect the plant growth.

(b) Gary wanted to make sure the size of the container did not affect differential plant growth in
his experiment.

(d) Gary wanted his garden to look organized.

(e) There is no possible scientific reason for having the same size containers.

8. What scientific reason might Gary have for insisting that all plants receive the same amount of water
everyday?

(a) Gary wanted to test the effect of shade on plant growth and therefore, he wanted to have no
variables other than the amount of sunshine on the plants.

(b) Gary wanted to test the effect of the amount of water on plant growth.

(d) Gary was conserving water.

(e) There is no possible scientific reason for having the same amount of water for each plant every
day.

9. What was the variable being tested in Gary's experiment?

(a) the amount of water

(b) the amount of plant food

(d) the amount of sunshine

(e) the type of soil

10. Which of the following factors may be varying in Gary's experimental setup that he did not control?

(a) individual plant variation

(b) soil temperature due to different colors of containers

(d) the effect of insects which may attack one set of plants but not the other

(e) All of the above are possible factors that Gary did not control

11. When a mosquito sucks blood from its host, it penetrates the skin with its sharp beak and injects
an anti-coagulant so the blood will not clot. It then sucks some blood and removes its beak. If the

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mosquito carries disease-causing microorganisms, it injects these into its host along with the anti-
coagulant. It was assumed for a long time that the virus of typhus was injected by the louse when
sucking blood in a manner similar to the mosquito. But apparently this is not so. The infection is
not in the saliva of the louse, but in the feces. The disease is thought to be spread when the louse
feces come in contact with scratches or bite wounds in the host's skin. A test of this was carried out
in 1922 when two workers fed infected lice on a monkey taking great care that no louse feces came
into contact with the monkey. After two weeks, the monkey had NOT become ill with typhus. The
workers then injected the monkey with typhus and it became ill within a few days. Why did the
workers inject the monkey with typhus near the end of the experiment?

(a) to prove that the lice carried the typhus virus

(b) to prove the monkey was similar to man

(d) to prove that mosquitoes were not carriers of typhus

(e) the workers were mean

12. Eijkman fed a group of chickens exclusively on rice whose seed coat had been removed (polished
rice or white rice). The chickens all developed polyneuritis (a disease of chickens) and died. He fed
another group of chickens unpolished rice (rice that still had its seed coat). Not a single one of them
contracted polyneuritis. He then gathered the polishings from rice (the seed coats that had been
removed) and fed the polishings to other chickens that were sick with polyneuritis. In a short time,
the birds all recovered. Eijkman had accurately traced the cause of polyneuritis to a faulty diet.
For the first time in history, a food deficiency disease had been produced and cured experimentally.
Which of the following is a reasonable statement of Eijkman's hypothesis?

(a) Polyneuritis is a fatal disease for chickens.

(b) White rice carries a virus for the disease polyneuritis.

(d) The rice seed coat contains a nutrient that provides protection for chickens against polyneuritis.

(e) None of these is a reasonable statement of Eijkman's hypothesis.

Questions 12, 13, and 14 relate to the following paragraphs.

Scientist A noticed that in a certain forest area, the only animals inhabiting the region were giraffes.
He also noticed that the only food available for the animals was on fairly tall trees and as the summer
progressed, the animals ate the leaves high and higher on the trees. The scientist suggested that these
animals were originally like all other animals but generations of animals stretching their necks to reach
higher up the trees for food, caused the species to grow very long necks.

Scientist B conducted experiments and observed that stretching muscles does NOT cause bones to grow
longer nor change the DNA of animals so that longer muscles would be passed on to the next generation.
Scientist B, therefore, discarded Scientist A's suggested answer as to why all the animals living in the area
had long necks. Scientist B suggested instead that originally many different types of animals including
giraffes had lived in the region but only the giraffes could survive when the only food was high in the trees,
and so all the other species had left the area.

12. Which of the following statements is an interpretation, rather than an observation?

A. The only animals living in the area were giraffes.

B. The only available food was on tall trees.

C. Animals which constantly stretch their necks will grow longer necks.

D. A, B, and C are all interpretations.

E. A, B, and C are all observations.

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13. Scientist A's hypothesis was that

A. the only animals living in the area were giraffes.

B. the only available food was on tall trees.

C. animals which constantly stretch their necks will grow longer necks.

D. the animals which possess the best characteristics for living in an area, will be the predominant species.

E. None of the above are reasonable statements of Scientist A's hypothesis.

14. Scientist A's hypothesis being discarded is

A. evidence that the scientific method doesn't always work.

B. a result achieved without use of the scientific method.

C: an example of what happened before the scientific method was invented.

D. an example of the normal functioning of the scientific method.

E. an unusual case.

15. When a theory has been known for a long time, it becomes a law.

A. True

B. False

16. During Pasteur's time, anthrax was a widespread and disastrous disease for livestock. Many people
whose livlihood was raising livestock lost large portions of their herds to this disease. Around 1876, a horse
doctor in eastern France named Louvrier, claimed to have intvented a cure for anthrax. The influential
men of the community supported Louvrier's claim to have cured hundreds of cows of anthrax. Pasteur
went to Louvrier's hometown to evaluate the cure. The cure was explained to Pasteur as a multi-step
process during which: 1) the cow was rubbed vigorously to make her as hot as possible; 2) long gashes
were cut into the cows skin and turpentine was poured into the cuts; 3) an inch-thick coating of cow
manure mixed with hot vinegar was plastered onto the cow and the cow was completely wrapped in a
cloth. Since some cows recover from anthrax with no treatment, performing the cure on a single cow would
not be conclusive, so Pasteur proposed an experiment to test Louvrier's cure. Four healthy cows were to
be injected with anthrax microbes, and after the cows became ill, Louvrier would pick two of the cows
(A and B) and perform his cure on them while the other two cows (C and D) would be left untreated.
The experiment was performed and after a few days, one of the untreated cows died and one of them got
better. Of the cows treated by Louvrier's cure, one cow died and one got better. In this experiment, what
was the purpose of infecting cows C and D?

A. So that Louvrier would have more than two cows to choose from.

B. To make sure the injection actually contained anthrax.

C. To serve as experimental controls (a comparison of treated to untreated cows).

D. To kill as many cows as possible.

17. A hypothesis is

A. a description of a consistent pattern in obervations.

B. an observation that remains constant.

C. a theory that has been proven.

D. a tentative explanation for a phenomenon.

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18. A scientific law is

A. a description of a consistent pattern in obervations.

B. an observation that remains constant.

C. a theory that has been proven.

D. a tentative explanation for a phenomenon.

19. A number of people became ill after eating oysters in a restaurant. Which of the following statements
is a hypothesis about this occcurence?

A. Everyone who ate oysters got sick.

B. People got sick whether the oysters they ate were raw or cooked.

C. Symptoms included nausea and dizziness.

D. The cook felt really bad about it.

E. Bacteria in the oysters may have caused the illness.

20. Which statement best describes the reason for using experimental controls?

A. Experimental controls eliminate the need for large sample sizes.

B. Experimental controls eliminate the need for statistical tests.

C. Experimental controls reduce the number of measurements needed.

D. Experimental controls allow comparison between groups that are different in only one independent

variable.

21. A student decides to set up an experiment to determine the relationship between the growth rate of
plants and the presence of detergent in the soil. He sets up 10 seed pots. In five of the seed pots, he
mixes a precise amount of detergent with the soil and the other five seed pots have no detergent in the
soil. The 5 seed pots with detergent are placed in the sun and the five seed pots with no detergent are
placed in the shade. All 10 seed pots receive the same amount of water and the same number and type of
seeds. He grows the plants for two months and charts the growth every two days. What is wrong with his
experiment?

A. The student has too few pots.

B. The student has two independent variables.

C. The student has two dependent variables.

D. The student has no experimental control on the soil.

22. A scientist plants two rows of corn for experimentation. She puts fertilizer on row 1 but does not put
fertilizer on row 2. Both rows receive the same amount of sun and water. She checks the growth of the
corn over the course of five months. What is acting as the control in this experiment?

A. Corn without fertilizer.

B. Corn with fertilizer.

C. Amount of water.

D. Height of corn plants.

23. If you have a control group for your experiment, which of the following is true?
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A. There can be more than one difference between the control group and the test group, but not more

three differences or else the experiment is invalid.

B. The control group and the test group may have many differences between them.

C. The control group must be identical to the test group except for one variable.

D. None of these are true.

24. If the hypothesis is rejected by the experiment, then:

A. the experiment may have been a success.

B. the experiment was a failure.

C. the experiment was poorly designed.

D. the experiment didn't follow the scientific method.

25. A well-substantiated explanation of an aspect of the natural world is a:

A. theory.

B. law.

C. hypothesis.

D. None of these.

Further Reading / Supplemental Links

• http : //learner . org/resources/series61 . html

The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos.
You are required to register before you can watch the videos but there is no charge. The website has two
videos that apply to this lesson. One is a video called The World of Chemistry that relates chemistry
to other sciences and daily life. Another video called Thinking Like Scientists relates to the scientific
method. The audience on the video is young children but the ideas are full grown.

• Website with lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 1-2 is on
the scientific method, http://www.fordhamprep.org/gcurran/sho/sho/lessons/lessonl2.htm

• Website of the James Randi Foundation. James Randi is a staunch opponent of fake science, http :
//www . randi . org/site/

• Websites dealing with the history of the scientific method, http : //www. historyguide . org/earlymod/
lecturelOc . html http : //www . history . boisestate . edu/WESTCIV/science/

Vocabulary

hypothesis A proposal intended to explain a set of observations.

theory A hypothesis that has been supported with repeated testing.

law A relationship that exists between specific observations.

experiment The act of conducting a controlled test or observations.

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scientific method A method of investigation involving observation to generate and test hypotheses and
theories.

superstition An irrational belief that an object, action, or circumstance not logically related to an event
influences its outcome.

Labs and Demonstrations for The Scientific Method
Teacher's Resource Page for Candle Observation

Investigation and Experimentation Objectives

In this activity, the student will be making and recording observations.

Safety Issues

If students are allowed to light their own candles, they should be instructed to strike matches on the
striker pad in a direction away from the body such that potential flying pieces of the burning match head
move away from the body. Extinguished matches should be held until cool, and then placed in solid
waste containers (wastebasket). Students should be reminded that during any lab involving an open flame
(candles, Bunsen burners, etc.) long hair must be restrained behind the head so that it does not fall
past the face when looking down. Students should be instructed not to handle candles once they are lit.
Dripping hot wax can be painful.

Observation List

The candle is cylinder in shape 1 and has a diameter 2 of about 2 cm. The length 3 of the candle was initially
about 18 cm and changed slowly during observation 4 , decreasing about 4 mm in 20 minutes.

The candle is made of a translucent, 5 white 6 solid 7 which has a slight odor 8 and no taste. 9 It is soft enough
to be scratched with a fingernail. 10 There is a wick which extends from the top to bottom of the candle
along its central axis 11 and protrudes 12 about 10 mm above the top of the candle. The wick is made of
three strands 13 of string braided 14 together.

The candle is lit by holding a source of flame close to the wick for a few seconds. 15 Thereafter, the source
of the flame can be removed and the flame sustains itself 16 at the wick.

The burning candle makes no sound. 17 While burning, the body of the candle remains cool to the touch 18
except near the top. 19 Within about 5 mm from the top of the candle, it is warm 20 but not hot, and
sufficiently soft to mold 21 easily.

The flame flickers 22 in response to air currents and tends to become smoky 23 while flickering. In the absence
of air currents, the flame is in the form shown in the picture 24 although it retains some movement 25 at all
times.

The flame begins 26 about 4 mm above the top of the candle, and at its base, the flame has a blue tint. 27
Immediately around the wick in a region about 5 mm wide 28 and extending about 8 mm above 29 the top
of the wick, the flame is dark. 30 This dark region is roughly conical 31 in shape.

Around this dark zone and extending about 5 mm above the dark zone is a region which emits yellow
light, 32 bright 33 but not blinding. The flame has rather sharply defined sides 34 but a ragged 35 top.

The wick is white 36 where it emerges from the candle, but from the base of the flame to the end of wick, it
is black, 37 appearing burnt except for the last 2 mm where it glows red. 38 The wick curls 39 over about 4 mm

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from its end. As the candle becomes shorter, the wick shortens 40 too, so as to extend roughly a constant
distance above the top of the candle.

Heat is emitted 41 by the flame, enough so that it becomes uncomfortable in a few seconds to hold ones'
fingers near the flame.

The top of a quietly burning candle becomes wet 42 with a colorless 43 liquid and becomes bowl-shaped. 44
Sometimes, the liquid in the bowl drains 45 down the side of the candle, cools, gradually solidifying 46 and
attaching itself to the candle. 47

Under quiet conditions, a stable 48 pool of clear liquid remains in the bowl-shaped top of the candle. The
liquid rises slightly around the wick, wetting 50 the base of the wick as high as the base 51 of the flame.

Candle Observation

Materials

Each student or pair of students is given a candle to observe. The candles should be between ^ an d 3 inch
in diameter (so their length will significantly shorten during the period) and mounted securely in a candle
holder (a jar lid will work fine). The teacher should light the candles, give appropriate instructions about
hair (it burns), and other safety issues. The student should be instructed to make as many observations
about the burning candle as they can in the allotted time. The teacher should encourage estimated
quantitative observations.

After the observation period, the teacher can ask for observations from the class and get as many as possible
on the board. Further observations can be added from the list above. It is useful for students to recognize
that there are many more observations about a simple system than they may have imagined.

Teacher's Resource Page for DAZOO

Investigation and Experimentation Objectives

In this activity, the student will use critical thinking to note evidence and logically draw a conclusion.

23 www.ckl2.org

Table 1.4: Answers for DAZOO

Question Number

Answer

Reason

4
5

7.
8
9
10

11.

12.

13.
14.
15.
16.

ZAM

NOOT

NIX

GOBBIE
BOBO

YATZ

BOBO

YATZ

CLINT

ARDZU

Girl

SLIP

BOBO
YATZ
WHEE
NIX

It is a family and they are outside
the cages.

It is a single male outside the
cages.

It is the largest number of indi-
viduals in a cage.
They have eight legs.
There is no large circle in the
cage.

There is no large rectangle in the
cage.

There are two small circles in the
cage.

There are three small rectangles
in the cage.

There are no small circles or rect-
angles in the cage.
Circle in circle - mother is preg-
nant - can't be SLIP because the
child is inside of a male.
The baby is a circle - which rep-
resents female.

There is a small BOBO rectangle
inside the large SLIP rectangle.
The crocodile ate a small giraffe.
They have wings.
They have no legs.
They live between YATZ and
CLINT.

Observation Game: DAZOO

SET-UP: Teacher prints out a copy of the game image and a set of questions for each student. With a
little thought, most students can answer the questions without teacher input.

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GOBBIE

BOBO

WHEE

ARDZU

DAZOO

NOOT

ZAM

YATZ

NIX

CLINT

SLIP

x n x

xVx

Only one family and the zookeeper are at the zoo today. The zookeeper is a single male, but the family
visiting the zoo has both males and females in the family. The family groupings of the zoo families have
been given surnames and are identified by the surnames in the diagram. Try to answer the questions below.

1 . What is the name of the family visiting the zoo?

2. What is the name of the zookeeper?

3. Which family in the zoo has the most members?

4. What is the name of the family of spiders?

5. Which mother is away at the hospital?

6. Which family has no father?

7. Which family has two daughters?

8. Which family has three sons?

9. Mr. and Mrs. Elephant have no children. Which family are they?
10. What will be the last name of the baby tiger when it is born?

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11. Will the baby tiger be a boy or a girl?

12. Mr. Crocodile has swallowed the giraffes son. Which is the family of crocodiles?

13. Which is the family of giraffes?

14. Which is the family of pelicans?

15. Which is the family of snakes?

16. The aardvarks live between the pelicans and elephants. Which is the family of aardvarks?

The Seven-of-Diamonds Game

Investigation and Experimentation Objectives

In this activity, the student will make and record observations, generate a hypothesis, and test the hy-
pothesis against further experimental observations. If the hypothesis fails, the student will add the new
observations to his list and create another hypothesis. If the hypothesis succeeds, the student "wins" the
game.

How to Play

This is the easiest of the observation and hypothesis games. To play the game, the teacher must select and
instruct an assistant to play the role of "psychic". The teacher draws the seven-of-diamonds set up on the
board as shown below.

!♦♦

♦ ♦:

\ ♦

u ♦

!♦ ♦

♦

♦ \

♦ ♦<

♦ ♦:

♦ ♦

To begin the game, the "psychic" is sent out of the room. While the psychic is out of the room, the
students select one of the cards and inform the teacher which card. Then the psychic is called back into
the room and the teacher points to one of the cards and asks the psychic, "Is this the card?" The psychic
responds either "yes" or "no" and the process continues until the teacher points to the correct card and the
psychic correctly identifies the card as the one the students had selected. The students are to observe the
game and after each trial run, make hypotheesis about how the trick is being done. They can do this as
individuals or in groups, the game can be played over and over until at least one student or group figures
out how the trick is being done. When a student or group thiks they know the trick, they can go out of
the room with the psychic and then play the role of the psychic when they return. If they can correctly

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identify the selected card, they win the game. The game can continue until more students figure it out or
the winners can explain the trick to those who didn't figure it out.

!♦♦

♦ ♦;

I ♦

!♦ ♦

♦♦♦

♦

♦ i

z t\

♦ ♦;

♦

♦ ♦

THE TRICK

The layout of the cards on the board and the spots on the seven-of-diamonds exactly correspond. The
psychic will not know the correct card until the teacher points to the seven-of-diamonds and asks, "Is this
the card?" When the teacher points to the seven-of-diamonds, he/she points to the spot on the card that
corresponds to the card selected. In the picture above, the teacher is pointing to the spot on the seven-
of-diamonds that corresponds to the position of the "6" in the layout. Therefore, the six-of-diamonds is
the selected card for this trial. The psychic continues to say "no" until the teacher points to the six-of-
diamonds and then says "yes." You should note, it is not possible to have the psychic identify the correct
card on the first try unless the selected card is the seven-of-diamonds. If the seven is the selected card,
the teacher can point to the seven first and point to the center position. The psychic must be alert to get
this one. The teacher can vary the sequence of asking so that sometimes, the selected card is pointed to
on the second try or the fourth try, and so forth.

"This" or "That" Psychic Game

I i )

Investigation and Experimentation Objectives

How to Play

This is the most difficult of these observation/hypothesis games for the students to figure out. The teacher
draws 3 columns of 3 squares each on the board as shown at right. Once again, the teacher needs an
assistant to act as psychic. Secretly, the teacher and the "psychic" conspire and assign the two outside
columns to be called "this" columns and the middle column to be called a "that" column.

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As usual, the psychic leaves the room and the students select one of the squares to be "psychically"
identified. The psychic is called back into the room and the teacher proceeds to point at various squares
and ask the psychic, "Is it this one?, or "Is it that one?" The code known only to the teacher and the
psychic is that if the teacher uses the correct name of the column when inquiring about a square, the
psychic answers "no." If the teacher uses the incorrect name of the column when inquiring, the psychic
replies "yes."

□ □□

This Column That Column This Column

Is it this one? (Correct title) NO

Ts it this one? (Correct title) NO

Is it this one? (Incorrect title) YES

One of the things that make this game so difficult is that the teacher can ask about the correct square on
the first try. On the very first trial, the teacher can point to a square in the middle column (the "that"

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column) and ask, "Is it this one?" and the psychic replies "yes."

1.2 Chemistry in History

Lesson Objectives

Give a brief history of how chemistry began.

State the Law of Conservation of Mass.

Explain the concept of a model, and create simple models from observations.

Introduction

During medieval times, a group of people known as alchemists began looking for ways to transform common
metals, such as lead, copper and iron, into gold (Figure ??). Can you imagine how much money you would
make if you could go to the store, buy some iron nails, and turn them into gold? You'd be rich in no time!

Alchemists experimented with many different kinds of chemicals, searching for what they termed the
"philosopher's stone" - a legendary substance that was necessary for the transformation of common metals
into gold. We now know that there is no such thing as a "philosopher's stone," nor is there any chemical
reaction that creates gold from another metal. We know this because we now have a much better under-
standing of the matter in our universe. Nevertheless, it was thanks to those early alchemists that people
became interested in chemistry in the first place.

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"Chemistry" was Derived from an Arabic Word

When we speak of "chemistry," we refer to the modern, scientific study of matter and the changes that it
undergoes. Still, it's no coincidence that the word "chemistry," looks a lot like the word "alchemy." Early
alchemists were commonly known as 'chemists,' and over time, people started referring to their work,
particularly the more legitimate forms of it, as chemistry. In many ways, it's appropriate that our word for
the present-day study of matter comes from the early practice of alchemy because a lot of the techniques
and equipment fundamental to modern chemistry were actually developed by early alchemists.

The origin of the word "alchemy" is something of a mystery. Certainly, early Europeans borrowed
"alchemy" from the Arabic word "al-kimia, " meaning "the art of transformation" (of course, the trans-
formation that alchemists were primarily concerned with involved the creation of gold). Most of what we
know today about early alchemy is based on translations of Arabic documents. That's because Muslim
alchemists were some of the first to keep careful notes about their experiments.

Even though our earliest records of alchemy come from the Arab Empire, some scholars believe that Arabs
adopted alchemy and the word "al-kimia" from the Greeks around 650 AD. The Greeks, in turn, may have
learned of alchemy from the Egyptians. Khem was an ancient name for Egypt, and Egyptians were known,
in early history, as masters of the art of working with gold. It's very likely that "al-kimia" is actually a
distorted version of the word "al-kimiya," meaning "the art of the land of Khem," or the art of Egypt.

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The Origins of Chemistry was Multi- Cultural

While the word "chemistry" may have its roots in Egypt, chemical experimentation seems to have been
prevalent all over the world, even as early as the 5 century BC. In China, the goal of the early alchemists
was largely to find "the elixir of life" - a potion that could cure all diseases and prevent death. Ironically,
many of these early elixirs involved mixtures of mercury and arsenic salts, both of which are extremely
poisonous. In fact, it's rumored that several Chinese emperors actually died after drinking "elixirs of life"
(Figure 1.10). Despite never finding a magical potion that could cure all diseases, early Chinese chemists
did discover many new chemicals and chemical reactions, including those used in fireworks and gunpowder.

Figure 1.10: Jiajing Emperor, rumored to have died after drinking poison he believed to be "the elixir of
life."

Like the Chinese, alchemists from India were interested in some of the medical benefits of different chem-
icals. In addition to medicine, Indian alchemists were fascinated by metals and metallurgy. Early Indian
writings contain methods for extracting and purifying metals like silver, gold and tin from ores that were
mined out of the ground. Moreover, it was alchemists from the Indian subcontinent who first realized that
by mixing molten metals with other chemicals they could produce materials that had new and beneficial
properties. For example, Wootz steel (also known as Damascus steel) was a substance discovered in Sri
Lanka around 300 AD. It was made by mixing just the right amounts of molten iron, glass and charcoal, but
it became famous because it could be used to produce swords, legendary for their sharpness and strength
in battle.

Around the same time that Wootz steel was being developed in Sri Lanka, Egyptian and Greek alchemists
were beginning to experiment as well. Much of the alchemy in this part of the world involved work
with colors and dyes and, of course, the transformation of common metals into gold and silver. Greek
philosophers like Plato and Aristotle, however, were also responsible for the important suggestion that

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the universe could be explained by unified natural laws. As you will discover in the next section, modern
chemistry relies on the study of these universal laws. Of course, modern scientific laws are slightly different
than the laws that either Plato or Aristotle had in mind. In general, scientific laws are determined by
careful experimentation and observation, whereas the early Greeks believed that their "natural laws"
could be deduced through philosophy.

Medieval Europeans were similarly fascinated by alchemy. Unfortunately, many alchemists in Europe
borrowed ideas from the more mystical of the Arabian alchemists and, as a result, European alchemy
quickly became associated with wizardry, magic, and the search for the "philosopher's stone." It wasn't
until the late 17' A century that European chemists began applying the scientific method. Robert Boyle
(1627 - 1691) was the first European to do so, using quantitative experiments to measure the relationship
between the pressure and the volume of a gas. His use of the scientific method paved the way for other
European scientists and helped to establish the modern science of chemistry.

Figure 1.11: Antoine Lavoisier, "The Father of Modern Chemistry."

About 100 years after Robert Boyle first performed his experiments, a French scientist by the name of
Antoine Lavoisier (1743 - 1794) employed the scientific method when he carefully measured the masses
of reactants and products before and after chemical reactions (Figure 1.11). Since the total mass (or
quantity of material) never changed, Lavoisier's experiments led him to the conclusion that mass is neither
created nor destroyed. This is known as the Law of Conservation of Mass. Lavoisier is often called
"The Father of Modern Chemistry" because of his important contribution to the study of matter.

After the success of Lavoisier's work, experiments involving careful measurement and observation became
increasingly popular, leading to a rapid improvement in our understanding of chemicals and chemical
changes. In fact, by the end of the 19 century, chemical knowledge had increased so much that practically
everyone had stopped searching for the "philosopher's stone."

What Chemists Do

You might wonder why the study of chemistry is so important if you can't use it to turn iron into gold or
to develop a potion that will make you immortal. Why didn't chemistry die when scientists like Boyle and
Lavoisier proved alchemy was nothing but a hoax? Well, even though we can't use chemistry to make gold

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or to live forever, modern chemistry is still very powerful! There may be no such thing as a potion that
cures all diseases, but many chemists today are developing cures for specific diseases. In fact, chemists are
working on everything from treatments for HIV/AIDS to medications for fighting cancer.

Modern chemists study not only chemicals that can help us, but also chemicals that can hurt us. For
example, environmental chemists test the air, soil, and water in our neighborhoods to make sure that
we aren't exposed to heavy metals (such as mercury or lead) or chemical pesticides. Moreover, when
environmental chemists do find dangerous substances, they use their knowledge of chemistry to clean up
the contamination. Similarly, every time you buy packaged food from the grocery store, you can be sure
that many tests have been done by chemists to make sure that those foods don't contain any toxins or
carcinogens (cancer-causing chemicals).

Chemists are also responsible for creating many important materials we use today. Other technologies
rely on chemistry as well. In fact, your flat-screen LCD TV, the cubic zirconium ring on your finger, and
the energy efficient LED lights in your home are all thanks to our improved understanding of chemistry
(Figure 1.12).

Figure 1.12: Energy efficient LED lights can be used to brighten your home for a party or a holiday.

So, how do chemists accomplish all of these remarkable achievements? Unlike many of the early alchemists
that experimented by randomly mixing together anything that they could find, today's chemists use the
scientific method. This means that chemists rely on both careful observation and well-known physical laws.
By putting observations and laws together, chemists develop what they term models. Models are really
just ways of predicting what will happen given a certain set of circumstances. Sometimes these models are
mathematical, but other times, they are purely descriptive.

A model is any simulation, substitute, or stand-in for what you are actually studying. A good model con-
tains the essential variables that you are concerned with in the real system, explains all the observations
on the real system, and is as simple as possible. A model may be as uncomplicated as a sphere represent-
ing the earth or billiard balls representing gaseous molecules, or as complex as mathematical equations
representing light.

The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos.
You are required to register before you can watch the videos but there is no charge. After you register the
first time, you can return to the website (from the same computer) and view videos without registering
again. The website has a video that applies to this lesson. The video is called "Modeling The Unseen".
Video on Demand - Modeling the Unseen (http : //www. learner .org/resources/series61 .html?pop=
yes&pid=793#)

Over time, scientists have used many different models to represent atoms. As our knowledge of the atom
changed, so did the models we use for them. Our model of the atom has progressed from an "indestructible
sphere," to a dish of "plum pudding," to a "nuclear model."

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Hypotheses and theories comprise some ideas that scientists have about how nature works but of which
they are not completely sure. These hypothesis and theories are models of nature used for explaining and
testing scientific ideas.

Chemists make up models about what happens when different chemicals are mixed together, or heated up,
or cooled down, or compressed. Chemists invent these models using many observations from experiments
in the past, and they use these models to predict what might happen during experiments in the future.
Once chemists have models that predict the outcome of experiments reasonably well, those working models
can help to tell them what they need to do to achieve a certain desired result. That result might be the
production of an especially strong plastic, or it might be the detection of a toxin when it's present in your
food.

Science is not the only profession whose members make use of the scientific method. The process of making
observations, suggesting hypotheses, and testing the hypotheses by experiment is also a common procedure
for detectives and physicians.

Lesson Summary

• The word "chemistry" comes from the Arabic word "al-kimia" meaning "the art of transformation."

• Chemistry began as the study of alchemy. Most alchemists were searching for the "philosopher's
stone," a fabled substance that could turn common metals into gold.

• Chinese alchemists were particularly interested in finding "the elixir of life."

• In India, much early chemistry focused on metals.

• The scientific method involves making careful observations and measurements and then using these
measurements to propose hypotheses (ideas) that can, in turn, be tested with more experiments.

• Robert Boyle and Antoine Lavoisier employed the "scientific method," thereby bringing about the
rise of modern chemistry.

• The Law of Conservation of Mass states that mass is neither created nor destroyed.

• Modern chemists perform experiments and use their observations to develop models. Models then
help chemists to understand and predict the results of future experiments. Models also help chemists
to design new materials and cures for diseases.

Review Questions

1. Where does the word "chemistry" come from?

2. Consider the following data about John's study habits, and grades:

(a) Propose a qualitative (words, but no math) model that might describe how the length of time
John spends studying relates to how well he does on the test?

(b) If John wants to earn 92% on his next test, should he study for about 6 hours, 9 hours, 12 hours,
or 18 hours? Justify your answer.

Table 1.5:

Hours spent studying Grade earned on the test

20%

5 40%

10 60%

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Table 1.5: (continued)

Hours spent studying Grade earned on the test

3. Helen wanted to know if lemon juice chemically reacts with tea to lighten its color. So Helen added
25 drops of lemon juice to 250 mL of tea and observed that the tea colored lightened significantly.
Helen wanted to make sure that the color lightening was the result of a chemical reaction and not the
result of dilution. Which one of the following activities should Helen carry out to serve as a control
for this experiment?

(a) Helen should add 25 drops of orange juice to another 250 mL sample of tea.

(b) Helen should add 25 drops of distilled water to another 250 mL sample of tea.

(d) Helen should add 25 drops of tea to a 250 mL sample of lemon juice.

(e) Helen should add 25 drops of tea to a 250 mL sample of tea.

Vocabulary

hypothesis A proposal intended to explain a set of observations.

theory A hypothesis that has been supported with repeated testing.

law A relationship that exists between specific observations.

scientific method A method of investigation involving observation to generate and test hypotheses and
theories.

chemistry The science of the composition, structure, properties, and reactions of matter.

1.3 Chemistry is a Science of Materials

Lesson Objectives

• Give examples of chemical properties a scientist might measure or observe in a laboratory.

• Explain the difference between a physical change and a chemical change, giving examples of each.

• Identify the situations in which mass can be converted to energy and energy can be converted to
mass.

Introduction

In the last chapter we discussed some of the goals of early alchemists and some of the roles of chemists
today. What you might have noticed is that while methods of chemical experimentation have improved
and while knowledge of chemical properties has increased, chemistry in the 21 if century AD and chemistry
in the 5 th century BC were both concerned with the question: How does matter change from one form to
another? Can we predict the properties of matter? And how can we control these properties in order to
use them to our advantage? Chemistry is essentially concerned with the science of matter and materials.
Therefore, we'll begin our discussion of chemistry by considering some of the chemical materials that have
been important both to early civilizations and to society today.

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Ancient Materials Versus Modern Materials

Before humans had any understanding of chemistry, they used whatever they could find in the world around
them. One type of material that was easily accessible to early civilizations, at least in small amounts, was
metal. Native gold, silver, and copper, and compounds of tin and iron can all be found occurring naturally
in cliffs and caves (in fact, the discovery of natural gold in El Dorado County, California is what lead to
the great Gold Rush of 1849) and, as a result, these metals became very important to people in early times
(Figure ??).

Many ancient civilizations fashioned tools, jewelry, and weapons out of metal that they scavenged from rocks
around them (Figure ??). After a while, however, people discovered that by mixing naturally occurring
metals with other substances, they could create new materials that often had superior properties.

Some of the oldest materials produced by man include mixtures (or more specifically solutions) of metals
known as alloys. One of the earliest alloys ever discovered was bronze. Bronze can be made by heating
chunks of tin and copper until they are liquid and then mixing the two pure metals together. Bronze was
very important to early civilizations because it was more resistant to rust than iron, harder than copper,
and could hold an edge and be sharpened to create tools and weapons.

Another alloy, produced early in the history of civilization, is steel. As you learned in an earlier section,
steel is an alloy of iron and carbon (or charcoal). Steel, particularly Wootz steel (which required a special
technique that involved the addition of glass), was especially strong, and could be fashioned into very
sharp edges, perfect for swords. Another old material whose production was known to early civilizations
is brass. Again, brass is an alloy, made of two pure metals, copper, and zinc. Early Romans knew that if
they melted copper, and a zinc ore known as calamine together, they could produce brass, which was both

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shiny like gold, and resistant to rust. Brass was a common material used to make coins.

What you might notice about these "old" materials is that they are mainly alloys. At the time when bronze
and brass and steel were discovered, people didn't know much about the composition of matter or about
how matter was assembled on a microscopic scale. As a result, inventing materials was largely a matter of
trial and error. Towards the end of the 19 century, however, scientists were beginning to understand the
make-up of matter, and this understanding led to new insight into how to develop materials with desirable
properties.

8 JL*

One of the huge breakthroughs in recent history has been the discovery of plastic and plastic products
(Figure ??). Initially, plastic was made by chemically modifying cellulose, a naturally occurring chemical
found in plants. As chemical knowledge developed, however, scientists began to realize that plastics had
special properties because, on a microscopic scale, they were composed of thousands of tiny chains of
molecules all tangled up together. Scientists reasoned that if they altered the chemicals in these chains,
but still managed to keep the chains intact, they could make new plastics with new properties. Thus began
the plastic revolution!

Semiconductors are another class of "new" materials whose development is largely based on our improved
understanding of chemistry. Because scientists know how matter is put together, they can predict how to
fine-tune the chemical composition of a semiconductor in order to make it absorb light and act as a solar
cell or emit light and act as a light source. We've come a long way from our early days of producing bronze
and steel. Nevertheless, as our understanding of chemistry improves, we will be able to create even more
useful materials than we have today.

Chemists Study the Properties of Matter

Hopefully at this point you are fully convinced of how important and useful the study of chemistry can
be. You may, however, still be wondering exactly what it is that a chemist does. Chemistry is the study of
matter and the changes that matter undergoes. In general, chemists are interested in both characteristics
that you can test and observe, like a chemical's smell or color, and characteristics that are far too small to
see, like what the oxygen you breathe in or the carbon dioxide you breath out looks like under a microscope
1,000 times more powerful than any existing in the world today.

Wait a minute... how can a chemist know what oxygen and carbon dioxide look like under a microscope
that doesn't even exist? What happened to the scientific method? What happened to relying on observa-
tions and careful measurements? In fact, because chemists can't see the underlying structure of different

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materials, they have to rely on the scientific method even more! Chemists are a lot like detectives (Figure
??). Suppose a detective is trying to solve a murder case - what does she do? Obviously, the detective
starts by visiting the site of the crime and looking for evidence. If the murderer has left enough clues
behind, the detective can piece together a theory explaining what happened.

Even though the detective wasn't at the crime scene when the crime was committed and even though
the detective didn't actually see the murderer kill the victim, with the right evidence, the detective can
be pretty sure he or she knows how it took place. It's the same with chemistry. When chemists go into
the laboratory, they collect evidence by making measurements. Once they've collected enough clues from
the properties that they can observe, they use that evidence to piece together a theory explaining the
properties that they can't observe - the properties that are too small to see.

What kinds of properties do chemists actually measure in the laboratory? Well, you can probably guess a
few. Imagine that you go to dinner at a friend's house and are served something that you don't recognize -
what types of observations might you make to determine exactly what you've been given? You might smell
the food. You might note the color of the food. You might try to decide whether the food is a liquid or a
solid because if it's a liquid, it's probably soup or a drink. The temperature of the food could be useful if
you wanted to know whether or not you'd been served ice cream! You could also pick up a small amount
of food with your fork and try to figure out how much it weighs - a light dessert might be something like
an angel cake, while a heavy dessert is probably a pound cake. The quantity of food you've been given
might be a clue too. Finally, you might want to know something about the food's texture - is it hard and
granular like sugar cubes, or soft and easy to spread, like butter?

Believe it or not, the observations you are likely to make when trying to identify an unknown food are very
similar to the observations that a chemists makes when trying to learn about a new material. Chemists
rely on smell, color, state (that is, whether it is a solid, liquid or gas), temperature, volume, mass (which
is related to weight, as you'll discover in a later section), and texture. There is, however, one property you
might use to learn about a food, but that you should definitely not use to learn about a chemical - taste!

In The Atomic Theory, you'll see exactly how measurements of certain properties helped early scientists to
develop theories about the chemical structure of matter on a scale much smaller than they could ever hope
to see. You'll also learn how these theories, in turn, allow us to make predictions about new materials that
we haven't even created yet.

The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos.
You are required to register before you can watch the videos but there is no charge. After you register the
first time, you can return to the website (from the same computer) and view videos without registering
again. The website has two videos that apply to this lesson. One is a video called The World of
Chemistry that relates chemistry to other sciences and daily life. Another video called Thinking Like
Scientists relates to the scientific method. The audience on the video is young children but the ideas
are full grown. Video on Demand - The World of Chemistry (http://www.learner.org/resources/
series61.html?pop=yes&pid=793#)

Chemists Study of How and Why Matter Changes

In the last section, we discussed the properties of matter and how scientists use these properties to deduce
certain facts about the structure of matter. However, if chemists only studied properties such as color and
smell, they would only be collecting half of the evidence. While the properties of matter can tell us a lot,
so too can the changes that matter undergoes. Suppose you've been served a slice of cake that you've
noticed is cold to the touch (Figure ??). You might guess that you're dealing with an ice cream cake.
But then again, maybe it's just a normal cake that's been kept in the freezer. Can you think of some way
to tell between a frozen slice of ice cream cake and a frozen slice of regular cake? Well, one possibility is

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to wait for a while and see whether your slice melts. If the slice melts, it was an ice cream cake, and if it
doesn't, it was just regular cake. In this case, you aren't observing a property, but rather a change in a
property. The property being changed in the example is state.

Similarly, chemists learn a lot about the nature of matter by studying the changes that matter can undergo.
Chemists make a distinction between two different types of changes that they study - physical changes and
chemical changes. Physical changes are changes that do not alter the identity of a substance. Some
types of physical changes include:

• Changes of state (changes from a solid to a liquid or a gas and vice versa)

• Separation of a mixture

• Physical deformation (cutting, denting, stretching)

• Making solutions (special kinds of mixtures)

When you have a jar containing a mixture of pennies and nickels and you sort the mixture so that you
have one pile of pennies and another pile of nickels, you have not altered the identity of either the pennies
or the nickels - you've merely separated them into two groups. This would be an example of a physical
change. Similarly, if you have a piece of paper, you don't change it into something other than a piece of
paper by ripping it up. What was paper before you starting tearing is still paper when you're done. Again,
this is an example of a physical change (Figure ??).

You might find it a little harder to understand why changes in state are physical changes. Until we discuss
chemicals in terms of the smaller units (atoms and molecules) that make them up, it probably won't be
clear to you why freezing a substance or boiling a substance is only a physical change.

For now, though, you just have to trust that changes in state are physical changes. If you're ever in doubt,
remember this: when a lake freezes in the winter, the water doesn't disappear or turn into something else
- it just takes on a new form. Liquid water and solid water (ice) are just different forms of the substance
we know as water. For the most part, physical changes tend to be reversible - in other words, they can
occur in both directions. You can turn liquid water into solid water through cooling; you can also turn
solid water into liquid water through heating.

The other type of change that chemists are concerned with is chemical change. A chemical change occurs
when one substance is turned into an entirely new substance as a result of a chemical reaction (Figure ??).

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Again, as we learn more about chemicals, and what chemicals look like, the meaning of a chemical change
and the distinction between a chemical change and a physical change will become more obvious. For now,
realize that chemicals are made up of tiny units known as atoms. Some of these atoms are bonded (or
"glued") together, but during a chemical change, some of the bonds are broken and new bonds are formed.

You're probably wondering how you know when a chemical change has occurred. Sometimes it can be
pretty tricky to tell, but there are several evidences of chemical changes to look for. There has probably
been a chemical change if:

• A change in color has occurred

• Light, heat or sound has been given off from the material itself

• A precipitate (a solid formed when two liquids are mixed) has appeared

• A gas has been produced (detected by bubbling or a new odor)

Chemical changes are frequently harder to reverse than physical changes. One good example of a chemical
change is burning paper. In contrast to the act of ripping paper, the act of burning paper actually results in
the formation of new chemicals (carbon dioxide and water, to be exact). Notice that whereas ripped paper
can be at least partially reassembled, burned paper cannot be "unburned." In other words, burning only
goes in one direction. The fact that burning is not reversible is another good indication that it involves a
chemical change.

Chemists Study the Interchange of Matter and Energy

Chemists are concerned with the properties of matter, and the changes that matter undergoes. For the
most part, though, the changes that chemists are interested in are either physical changes, like changes in
state, or chemical changes like chemical reactions. In either case, Lavoisier's Law of Conservation of Mass,
applies. In both physical and chemical changes, matter is neither created nor destroyed. There is, however,
another type of change that matter can undergo that actually disobeys Lavoisier's Law of Conservation of
Mass, and that is the conversion of matter into energy, and vice versa.

Back when Lavoisier was studying chemistry, the technology didn't exist that would allow scientists to turn
matter into energy and energy into matter, but it can be done. This is the concept that Einstein proposed
in his famous equation E = mc 2 . (This equation states that the energy in a given amount of matter is equal
to the mass of the matter times the speed of light squared.) Chemical reactions don't involve changing
measureable amounts of energy to mass or mass to energy. Mass-energy conversion is, however, important
in chemistry that deals with radioactivity and particularly in the production of electricity by nuclear power
plants.

Lesson Summary

• Some of the earliest materials invented by humans were alloys such as bronze, steel, and brass.

• With improved understanding of chemistry comes the ability to design new and useful materials, like
plastics and semiconductors.

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• Chemists can't actually see the underlying structure of most materials. As a result, they measure
properties that they can see or observe and use this evidence to develop theories that explain how
chemicals are organized on a sub-microscopic (smaller than you can see with a microscope) scale.

• Some of the physical properties that scientists observe pertain to state (solid, liquid, or gas), tem-
perature, volume, mass, and texture.

• Chemists also study the changes that different materials undergo; this can give them valuable infor-
mation about the chemicals involved.

• There are two types of changes that are important in chemistry - physical changes and chemical
changes.

• Physical changes are changes that do not alter the identity of a substance; they are usually reversible.

• Chemical changes are changes that occur when one substance is turned into another substance as a
result of a chemical reaction. They are usually difficult to reverse.

• It is also possible to change matter into energy and energy into matter.

Review Questions

1. Name the two types of changes that chemists are primarily interested in.

2. Decide whether each of the following statements is true or false.

(a) Plastics were developed in Rome around 300 AD

(b) Bronze is an example of an alloy

(d) Brass is an example of an alloy

3. Match the following alloys with their common names.

(a) Bronze - a. tin and copper

(b) Brass - b. is not an alloy

(d) Tin - d. copper and zinc

(e) Steel - e. is not an alloy

4. Decide whether each of the following statements is true or false.

(a) Physical changes are typically accompanied by a color change

(b) A burning campfire is an example of a chemical change

(d) When you drop a plate, and it breaks, the plate undergoes a physical change

5. In each of the following examples, determine whether the change involved is a physical change or a
chemical change.

(a) Flattening a ball of silly putty

(b) Combining a bowl of cherries and a bowl of blueberries

(d) Cooking an egg

6. Judy has two beakers filled with clear liquids, and she needs to know whether the liquid in the
first beaker is the same as the liquid in the second beaker. In which scenario does Judy use physical
properties to answer her question, and in which scenario does Judy use changes in chemical properties
to answer her question?

(a) Judy smells the two liquids and notices that the liquid in the first beaker has a strong odor,
while she can't smell the liquid in the second beaker at all.

(b) Judy mixes some table salt into the first beaker and notices that a white precipitate forms. She
then mixes some table salt into the second beaker, but nothing happens.

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Vocabulary

alloy A solution (or a special kind of mixture), in which at least one of the components is a metal.

physical change Changes that do not alter the identity of the substance.

chemical change A change that occurs when one substance is turned into an entirely new substance as
a result of a chemical reaction.

1.4 Matter

Lesson Objectives

• Define matter and explain how it is composed of building blocks known as "atoms."

• Distinguish between mass and weight.

Introduction

We are all familiar with matter. The definition of matter is anything that has mass and volume (takes
up space). For most common objects that we deal with every day, it is fairly simple to demonstrate that
they have mass and take up space. You might be able to imagine, however, the difficulty for people several
hundred years ago to demonstrate that air has mass and volume. Air (and all other gases) are invisible
to the eye, have very small masses compared to equal amounts of solids and liquids, and are quite easy to
compress (change volume). Without sensitive equipment, it would have been difficult to convince people
that gases are matter. Today, we can measure the mass of a small balloon when it is deflated and then blow
it up, tie it off, and measure its mass again to detect the additional mass due to the air inside. The mass
of air, under room conditions, that occupies a one quart jar is approximately 0.0002 pounds. This small
amount of mass would have been difficult to measure in times before balances were designed to accurately
measure very small masses. Later, scientists were able to compress gases into such a small volume that
the gases turned into liquids, which made it clear that gases are matter.

On the other hand, when you add heat to an object, the temperature of the object increases, but even the
most sensitive balance cannot detect any difference in mass between an object when cold and when hot.
Heat does not qualify as matter.

The Material in the Universe

Knowing that planets, solar systems, and even galaxies are made out of matter doesn't bring us any closer
to understanding what matter is. Up until the early 1800s, people didn't really understand matter at all.
They knew that there were "things" in the world that they could pick up and use, and that some of these
"things" could be turned into other "things." For example, someone who found a piece of copper could
shape it into a necklace or melt it together with zinc to make brass. What people didn't know, though,
was how all of these "things" were related. If they had, the alchemists probably wouldn't have wasted so
much time trying to convert common metals into gold. You'll understand why by the time you're finished
with this section. Even though the universe consists of "things" as wildly different as ants and galaxies,
the matter that makes up all of these "things" is composed of a very limited number of building blocks
(Figure 1.13).

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Figure 1.13: Everything from an ant to an entire galaxy is composed of matter.

These building blocks are known as atoms, and so far, scientists have discovered or created a grand total
of 117 different types of atoms. Scientists have given a name to each different type of atom. A substance
that is composed of only one type of atom is called an element. Each element, therefore, has its own name;
it also has its own symbol. The "periodic table" is a way of summarizing all of the different atoms that
scientists have discovered (Figure 1.14). Each square in the periodic table contains the symbol (a capital
letter or a capital letter followed by a lower case letter) for one of the elements.

IA
1

alkali metals

post-transition metals [

alkaline earths

transition metals

1MB IVB VB VIB /MB VIII VIII
3 4 5 6 W 8 9

VIII IB\ MB
10 11 \l2

noble aases

IIIA
13

IVA VA VIA
14 15 16

VIIA
17 |

C N O

Si P S

VIIA

Li Be
Na Mg

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br
Cs Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn,Sb Te I

Rb Ba La
Fr Ra Ac

Hf Ta W Re Os Ir Pt Au

Bh Hs Mt Ds

Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

lanthanides

Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

antinides I

S.K. Lower

Figure 1.14: The Periodic Table.

At this point, what should amaze you is that all forms of matter in our universe are made with only
117 different building blocks. In some ways, it's sort of like cooking a gourmet, five-course meal using
only three ingredients! How is it possible? To answer that question, you have to understand the ways in
which different elements are put together to form matter. The most important method that nature uses
to organize atoms into matter is the formation of molecules. Molecules are groups of two or more atoms
that have been bonded together. There are millions of different ways to bond atoms together, which means
that there are millions of different possible molecules. Each of these molecules has its own set of chemical
properties, and it's these properties with which chemists are most concerned. You will learn a lot more
about atoms and molecules, including how they were discovered, in a later part of the textbook. Figure
1.15, however, gives you a preview of some of the common molecules that you might come in contact with
on a daily basis.

Now back to the question of why alchemists had trouble making gold out of other common metals. Most
naturally occurring metals, including gold, iron, copper, and silver are elements. Look carefully at the
Periodic Table in Figure 1.14. Do you see the symbol Au? Au is the symbol for gold. Gold is one of the

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oxygen

k 4

hydrogen
a water molecule

oxygen

CM)

carbon
a carbon dioxide molecule

a nitrogen dioxide molecule

nitrogen

hydrogen
an ammonia molecule

oxygen

an oxjgen molecule

Figure 1.15: Some common molecules and the atoms that they are made from.

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117 different types of atoms - it's one of nature's building blocks. Take another look at the Periodic Table.
Do you see the symbol Fe? Fe is the symbol for iron. Again, that means that iron is an element, or a type
of atom. In fact, copper (Cm), tin (Sn), and silver (Ag) are all elements. In other words, alchemists were
trying to convert one type of element, or building block, into another type of element, and that just can't
be done by chemical means.

Chemical reactions can turn elements into molecules, molecules into other molecules, and molecules back
into elements. Chemical reactions cannot, however, turn one type of element into another type of element.
The only way to do that is through what are known as nuclear reactions, and nuclear reactions require
advanced technical equipment that wasn't around in the days of the alchemists. It's like building a house.
You can make a house by cementing together bricks, stones, and wood, just like you can make a molecule
by bonding together different types of atoms. You can also get your bricks, stones, and wood back by
taking the house apart just like you can get your atoms back by taking the molecule apart. No matter how
hard you try, though, you can't turn bricks into wood. Converting common metals like copper into gold
or iron into gold would be like turning bricks into wood - it's simply not possible.

Matter Has Mass and Occupies Space

So far we've decided that the entire universe is composed of matter, which is in turn composed of atoms.
Frequently, though, chemists want to know how much matter they actually have. To figure this out they
rely on two fundamental properties of matter. All matter in the universe, from a teaspoon of salt to the
Pacific Ocean, has mass and occupies space. When scientists measure how much space is taken up by a
certain quantity of matter, they are measuring the objects volume. Obviously, the volume of the Pacific
Ocean is a lot larger than the volume of a teaspoon of salt. Unfortunately, while volume is an important
property, and plays an important role in a lot of different chemical experiments, volume is not the best
way to determine how much matter you have.

Typically we think that the bigger something is, the more there is in it. That's certainly true a lot of
the time in our everyday lives. If you pour yourself two cups of coffee in the morning, you'll be drinking
twice as much coffee as you would have if you'd only poured yourself a single cup. Unfortunately, any time
that we compare volumes in this way, we are making two assumptions that aren't always true in chemical
experiments. First, we are assuming constant temperature. That's important, because the amount of space
taken up by a certain quantity of matter depends on the temperature of that matter. In general, heating
something up causes it to expand, and cooling something down causes it to contract. Secondly, when you
compare volumes in everyday life, you are almost always comparing volumes of the same material. You
can compare two cups of coffee to one cup of coffee, but how do you compare two cups of coffee to one cup
of ice cream? It really doesn't make sense. Volume is not a good way to determine the quantity of matter
that you have.

If you can't use volume to figure out how much matter you've got, what can you use? It turns out that
the best way to determine quantities of matter is to use a measure known as mass. The mass of an object
doesn't change with temperature, which makes it a lot easier to determine how much stuff you're dealing
with, especially when you don't know what the temperature is or when the temperature keeps changing.
Another good thing about mass is that an atom of a particular element always has the same mass (Strictly
speaking that's not entirely true because of what are known as isotopes, but you won't need to worry about
that now). For example, an atom of gold always has a mass of about 197 atomic mass units or daltons.
Atomic mass units are units that we use to measure mass, just like a mile is a unit that we use to measure
distance, and an hour is a unit that we use to measure time. Even when atoms are bonded together into
molecules, the individual atoms have the same mass, meaning that by adding up all of the masses of the
atoms in a molecule, it's fairly easy to figure out the mass of the molecule itself. You'll eventually learn
how to do this.

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The Difference Between Mass and Weight

One typical mistake that students make when learning about mass for the first time is confusing mass with
weight. Again, this confusion is largely due to the fact that, in everyday English, we frequently use the
word "weight" when we actually mean "mass." For example, when you say "I want to lose weight," what
you really mean is, "I want to lose mass." In science, the word "weight" has a very specific definition that
is different from what you might expect. Do not confuse the everyday meaning of the word "weight" with
the scientific meanings of the word "weight" (Figure 1.16).

Figure 1.16: The type of balance a chemist might use to measure the mass of an object.

In science, mass is an intrinsic ("built-in") property of matter. The mass of an atom is the same regardless
of the temperature or the other atoms that are bonded to it.

Figure 1.17: With no large mass nearby, this spaceship is weightless.

Similarly, the mass of an atom doesn't change depending on where it is. The mass of an atom is the same
on Earth as it would be on the moon, or on Jupiter, or in the middle of space. Weight, on the other hand,
does change with location. An object that weighs 240 pounds on earth would weigh about 40 pounds on
the moon and that same object in a spaceship far away from any large mass would weigh zero (Figure
1.17). In all three cases, however, the object would have exactly the same mass.

In science, weight is a measurement of how strongly gravity pulls on an object. Weight depends on both
the mass of the object and the force of gravity the object is experiencing. That's why your weight changes
depending on where you're standing. In each case, your mass will be the same, but the force of gravity
on Earth is different than the force of gravity on the moon, or on Jupiter and, as a result, your weight is
different too. The force of gravity, however, doesn't change significantly on the surface of Earth. In other

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words, the force of gravity in California is approximately the same as the force of gravity in Australia. As
a result, as long as you stick to the surface of the Earth, the more massive an object is, the more it weighs.

Lesson Summary

• All physical objects are made of matter.

• Matter itself is composed of tiny building blocks known as "atoms." There are only 117 different types
of atoms known to man.

• Frequently, atoms are bonded together to form "molecules."

• All matter has mass and occupies space.

• Volume is a measure of how much space an object occupies. Volume is not a good measure of how
much matter makes up any given object.

• Mass is an intrinsic property of matter that does not depend on temperature, location, or the way in
which the matter is organized (how the atoms are bonded) As a result, mass is an excellent measure
of how much matter is in any given object.

• "Mass" and "weight" have two very different scientific meanings.

• "Mass" only depends on how much matter is in an object. "Weight," on the other hand, depends on
how strongly gravity pulls on an object.

Review Questions

1. What is matter?

2. In this chapter, we'll learn about atoms, which are the building blocks of all matter in the universe.
As of 2007, scientists only know of 117 different types of atoms. How do you think it's possible to
generate so many different forms of matter using only 117 types of building blocks?

3. Which do you think has more matter, a cup of water or a cup of mercury? Explain.

4. Decide whether each of the following statements is true or false.

(a) Mass and weight are two words for the same concept.

(b) Molecules are bonded together to form atoms.

(d) The symbol for Gold in the periodic table is Gd.

5. Would you have more mass on the moon or on Earth?

6. Would you have more weight on the moon or on Earth? The force of gravity is stronger on the Earth
than it is on the moon.

7. Match the following terms with their meaning.

(a) Mass - a. a measure of the total quantity of matter in an object

(b) Volume - b. a measure of how strongly gravity pulls on an object

8. For the following statements, circle all of the options that apply: Mass depends on...

(a) the total quantity of matter

(b) the temperature

(d) the force of gravity

Volume depends on...

(a) the total quantity of matter

(b) the temperature

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(d) the object's size (independent of shape)

Weight depends on...

(a) the total quantity of matter

(b) the temperature

(d) the force of gravity

Further Reading / Supplemental Links

Website with lessons, worksheets, and quizzes on various high school chemistry topics.

• Lesson 1-4 is on the Classification of Matter.

• Lesson 1-5 is on Physical and Chemical Properties and Changes, http://www.fordhamprep.org/
gcurran/sho/sho/lessons/lessonl4 . htm

Vocabulary

matter Anything of substance that has mass and occupies space.

atom The basic building block of all matter. There are 117 known types of atoms. While atoms can be
broken down into particles known as electrons, protons and neutrons, this is very difficult to do.

element A type of atom. There are 117 known elements.

molecule Two or more atoms bonded together. Specific molecules, like water, have distinct characteris-
tics.

Periodic Table A way of summarizing all the different atoms that scientists have discovered. Each
square in the periodic table contains the symbol for one of the elements.

mass An intrinsic property of matter that can be used to measure the quantity of matter present in a
sample.

weight A measurement of how strongly gravity pulls on an object.

volume A measurement of how much space a substance occupies.

1.5 Energy

Lesson Objectives

• Define heat and work.

• Distinguish between kinetic energy and potential energy.

• State the law of conservation of matter and energy.

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Introduction

Just like matter, energy is something that we are all familiar with and use on a daily basis. Before you
go on a long hike, you eat an energy bar; every month, the energy bill is paid; on TV, politicians argue
about the energy crisis. But have you ever wondered what energy really is? When you plug a lamp into an
electric socket, you see energy in the form of light, but when you plug a heating pad into that same socket,
you only feel warmth. When you eat a bowl of spaghetti, the energy it provides helps you to function
throughout the day, but when you eat five bowls of spaghetti, some of that energy is turned into body fat.

If you stop to think about it, energy is very complicated. Still, we use energy for every single thing that
we do, from the moment we wake up to the moment we go to sleep. Without energy, we couldn't turn on
lights, we couldn't brush our teeth, we couldn't make our lunch, and we couldn't travel to school. Although
we all use energy, very few of us understand how we use it.

Ability to Do Work or Produce Heat

When we speak of using energy, we are really referring to transferring energy from one place to another.
When you use energy to throw a ball, you transfer energy from your body to the ball, and this causes
the ball to fly through the air. When you use energy to warm your house, you transfer energy from the
furnace to the air in your home, and this causes the temperature in your house to rise. Although energy is
used in many kinds of different situations, all of these uses rely on energy being transferred in one of two
ways. Energy can be transferred as heat or as work. Unfortunately, both "heat" and "work" are common
words, so you might think that you already know their meanings. In science, the words "heat" and "work"
have very specific definitions that are different from what you might expect. Do not confuse the everyday
meanings of the words "heat" and "work" with the scientific meanings.

When scientists speak of heat, they are referring to energy that is transferred from an object with a higher
temperature to an object with a lower temperature as a result of the temperature difference. Heat will
"flow" from the hot object to the cold object until both end up at the same temperature. When you cook
with a metal pot, you witness energy being transferred in the form of heat. Initially, only the stove element
is hot - the pot and the food inside the pot are cold. As a result, heat moves from the hot stove element to
the cold pot. After a while, enough heat is transferred from the stove to the pot, raising the temperature
of the pot and all of its contents (Figure 1.18).

Figure 1.18: Energy is transferred as heat from the hot stove element to the cooler pot until the pot and
its contents become just as hot as the element. The energy that is transferred into the pot as heat is then
used to cook the food.

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We've all observed heat moving from a hot object to a cold object, but you might wonder how the energy
actually travels. Whenever an object is hot, the molecules within the object are shaking and vibrating
vigorously. The hotter an object is, the more the molecules jiggle around. As you'll learn in the next
section, anything that is moving has energy, and the more it's moving, the more energy it has. Hot objects
have a lot of energy, and it's this energy that is transferred to the colder objects when the two come in
contact. The easiest way to visualize heat transfer is to imagine a domino effect.

Heat is being transferred from a hot object to a colder object. In detail: a. As the red molecules in the hot
object jiggle and vibrate, they hit some of the blue molecules in the colder object. This transfers energy
from the hot molecules to the colder molecules, causing these molecules to vibrate faster, b. - d. Just
like dominoes, heat passes along the chain until the energy is spread equally between all of the molecules.
(Source: Sharon Bewick. CC-BY-SA)

hot object

cold object

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Take a close look at the figure above. When the vibrating molecules of the hot object bump into the
molecules of the colder object, they transfer some of their energy, causing the molecules in the colder
object to start vibrating vigorously as well. As these molecules vibrate, they bump into their neighbors
and transfer some of their energy on down the chain. In this way, energy passes through the whole system
until all of the molecules have about the same amount, and the initial objects are at the same temperature.

Heat is only one way in which energy can be transferred. Energy can also be transferred as work. The
scientific definition of work is force (any push or pull) applied over a distance. Whenever you push an object
and cause it to move, you've done work, and you've transferred some of your energy to the object. At this
point, it's important to warn you of a common misconception. Sometimes we think that the amount of
work done can be measured by the amount of effort put in. This may be true in everyday life, but it isn't
true in science. By definition, scientific work requires that force be applied over a distance. It doesn't
matter how hard you push or how hard you pull. If you haven't moved the object, you haven't done any
work.

So far, we've talked about the two ways in which energy can be transferred from one place, or object, to
another. Energy can be transferred as heat, and energy can be transferred as work. But the question still
remains - what IS energy? We'll try to at least partially tackle that question in the next section.

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Types of Energy: Kinetic and Potential

Machines use energy, our bodies use energy, energy comes from the sun, energy comes from volcanoes,
energy causes forest fires, and energy helps us to grow food. With all these seemingly different types of
energy, it's hard to believe that there are really only two different forms of energy - kinetic energy and
potential energy. Kinetic energy is energy associated with motion. When an object is moving, it has
kinetic energy. When the object stops moving, it has no kinetic energy. While all moving objects have
kinetic energy, not all moving objects have the same amount of kinetic energy. The amount of kinetic
energy possessed by an object is determined by its mass, and its speed. The heavier an object is and the
faster it is moving, the more kinetic energy it has.

Kinetic energy is very common, and it's easy to spot examples of it in the world around you. Sometimes
we even try to capture kinetic energy and use it to power things like our home appliances. If you're from
California, you might have driven through the Tehachapi Pass near Mojave or the Montezuma Hills in
Solano County and seen the windmills lining the slopes of the mountains (Figure 1.19). These are two of
the larger wind farms in North America. As wind rushes along the hills, the kinetic energy of the blowing
air particles turns the windmills, trapping the wind's kinetic energy so that people can use it in their houses
and offices.

Figure 1.19: A wind farm in the Tehachapi Mountains of Southern California. Kinetic energy from the
rushing air particles turns the windmills, allowing us to capture the wind's kinetic energy and use it.

Capturing kinetic energy can be very effective, but if you think carefully, you'll realize that there's a small
problem. Kinetic energy is only available when something is moving. When the wind is blowing, we can
use its kinetic energy, but when the wind stops blowing, there's no kinetic energy available. Imagine what
it would be like trying to power your television set using the wind's kinetic energy. You could turn on the
TV and watch your favorite program on a windy day, but every time the wind stopped blowing, your TV
screen would flicker off because it would run out of energy. You'd probably only be able to watch about
half of the episodes, and you'd never know what was going on!

Of course, when you turn on the TV, or flip on the lights, you can usually count on them having a constant
supply of energy. This is largely because we don't rely on kinetic energy alone for power. Instead, we use
energy in its other form - we use potential energy. Potential energy is stored energy. It's energy that
remains available until we choose to use it. Think of a battery in a flashlight. If you leave a flashlight on,
the battery will run out of energy within a couple of hours, and your flashlight will die. If, however, you
only use the flashlight when you need it, and you turn it off when you don't, the battery will last for days
or even months. The battery contains a certain amount of energy, and it will power the flashlight for a
certain amount of time, but because the battery stores potential energy, you can choose to use the energy

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all at once, or you can save it and only use a small amount at a time.

Any stored energy is potential energy. Unfortunately, there are a lot of different ways in which energy
can be stored, and that can make potential energy very difficult to recognize. In general, an object has
potential energy because of its position relative to another object. For example when you hold a rock above
the earth, it has potential energy because of its position relative to the ground. You can tell that this is
potential energy because the energy is stored for as long as you hold the rock in the air. Once you drop
the rock, though, the stored energy is released.

There are other common examples of potential energy. A ball at the top of a hill stores potential energy
until it is allowed to roll to the bottom. When you hold two magnets next to each other, they store
potential energy too. For some examples of potential energy, though, it's harder to see how "position" is
involved. In chemistry, we are often interested in what is called chemical potential energy. Chemical
potential energy is energy stored in the atoms, molecules, and chemical bonds that make up matter. How
does this depend on position?

As you learned earlier, the world, and all of the chemicals in it are made up of atoms and molecules. These
store potential energy that is dependent on their positions relative to one another. Of course, you can't see
atoms and molecules. Nevertheless, scientists do know a lot about the ways in which atoms and molecules
interact, and this allows them to figure out how much potential energy is stored in a specific quantity (like
a cup or a gallon) of a particular chemical (Figure 1.20). Different chemicals have different amounts of
potential energy because they are made up of different atoms, and those atoms have different positions
relative to one another.

Two hydrogen atoms and an
oxygen atom making up a
water molecule

Figure 1.20: Scientists use their knowledge of what the atoms and molecules look like and how they interact
to determine the potential energy that can be stored in any particular chemical substance.

Since different chemicals have different amounts of potential energy, scientists will sometimes say potential
energy depends not only on position, but also on composition. Composition affects potential energy
because it determines which molecules and atoms end up next to each other. For example, the total
potential energy in a cup of pure water is different than the total potential energy in a cup of apple juice,
because the cup of water and the cup of apple juice are composed of different amounts of different chemicals.

At this point, you might be wondering just how useful chemical potential energy is. If you want to release
the potential energy stored in an object held above the ground, you just drop it. But how do you get
potential energy out of chemicals? It's actually not that difficult. You use the fact that different chemicals
have different amounts of potential energy. If you start with chemicals that have a lot of potential energy
and allow them to react and form chemicals with less potential energy, all the extra energy that was in the
chemicals at the beginning but not at the end is released.

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Law of Conservation of Matter and Energy

So far we've talked about how energy exists as either kinetic energy or potential energy and how energy
can be transferred as either heat or work. While it's important to understand the difference between
kinetic energy and potential energy and the difference between heat and work, the truth is, energy is
constantly changing. Kinetic energy is constantly being turned into potential energy, and potential energy
is constantly being turned into kinetic energy. Likewise, energy that is transferred as work might later end
up transferred as heat, while energy that is transferred as heat might later end up being used to do work.

Even though energy can change form, it must still follow one fundamental law - Energy cannot be created or
destroyed, it can only be changed from one form to another. This law is known as the Law of Conservation
of Energy. In a lot of ways, energy is like money. You can exchange quarters for dollar bills and dollar
bills for quarters, but no matter how often you convert between the two, you won't end up with any more
or any less money than you started with. Similarly, you can transfer (or spend) money using cash, or
transfer money using a credit card, but you still spend the same amount of money, and the store still
makes the same amount of money.

As it turns out, the law of conservation of energy isn't exactly the whole truth. If you think back, you'll
remember that energy and matter are actually interchangeable. In other words, energy can be created
(made out of matter) and destroyed (turned into matter). As a result, the law of conservation of energy
has been changed into the Law of Conservation of Matter and Energy. This law states that

The total amount of mass and energy in the universe is conserved (does not change).

This is one of the most important laws you will ever learn. Nevertheless, in chemistry we are rarely
concerned with converting matter to energy or energy to matter. Instead, chemists deal primarily with
converting one form of matter into another form of matter (through chemical reactions) and converting
one form of energy into another form of energy.

Let's take a look at several examples, where kinetic energy is switched to potential energy and vice versa.
Remember Wile E. Coyote with his anvil poised at the top of the cliff? As long as Wile E. Coyote holds
the anvil and waits for Road Runner, the anvil stores potential energy. However, when Wile E. Coyote
drops the anvil, the original potential energy stored in the anvil is converted to kinetic energy. The further
the anvil falls, the faster it falls, and more and more of the anvil's potential energy is converted to kinetic
energy.

The opposite, of course, happens when you throw a ball into the air. When the ball leaves your hand, it
has a lot of kinetic energy, but as it moves higher and higher into the sky, the kinetic energy is converted
to potential energy. Eventually, when all of the kinetic energy has been converted to potential energy, the
ball stops moving entirely and hangs in the air for a moment. Then the ball starts to fall back down, and
the potential energy is turned into kinetic energy again.

Just as kinetic energy and potential energy are interchangeable, work and heat are interchangeable too.
Think of a hot-air balloon (Figure 1.21). To operate a hot-air balloon, a flame at the base of the balloon
is used to transfer energy in the form of heat from the flame to the air molecules inside the balloon. The
whole point of this heat transfer, though, is to capture the heat and turn it into work that causes the
balloon to rise into the sky. The clever design of the hot-air balloon makes the conversion of heat to work
possible.

Lesson Summary

• Any time we use energy, we transfer energy from one object to another. Energy can be transferred
in one of two ways - as heat, or as work.

• Heat is the term given to energy that is transferred from a hot object to a cooler object due to the

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Figure 1.21: A hot air balloon transfers energy in the form of heat from the flame to the air particles in
the balloon. The design of the hot air balloon takes this energy and changes it from heat to work.

difference in their temperatures.

• Work is the term given to energy that is transferred as a result of a force applied over a distance.

• Energy comes in two fundamentally different forms - kinetic energy and potential energy.

• Kinetic energy is the energy of motion.

• Potential energy is stored energy that depends on the position of an object relative to another object.

• Chemical potential energy is a special type of potential energy that depends on the positions of
different atoms and molecules relative to one another. Chemical potential energy can also be thought
of as depending on chemical composition.

• Energy can be converted from one form to another.

• The total amount of mass and energy in the universe is conserved.

Review Questions

1 . Classify each of the following as energy primarily transferred as heat or energy primarily transferred
as work:

(a) The energy transferred from your body to a shopping cart as you push the shopping cart down
the aisle.

(b) The energy transferred from a wave to your board when you go surfing.

2. Decide whether each of the following statements is true or false:

(a) When heat is transferred to an object, the object cools down.

(b) Any time you raise the temperature of an object, you have done work.

(d) Any time you apply force to an object, you have done work.

3. Rank the following scenarios in order of increasing work:
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(a) You apply 100 N of force to a boulder and successfully move it by 2 m.

(b) You apply 100 N of force to a boulder and successfully move it by 1 m.

(d) You apply 200 N of force to a boulder but cannot move the boulder.

4. In science, a vacuum is denned as space that contains absolutely no matter (no molecules, no atoms,
etc.) Can energy be transferred as heat through a vacuum? Why or why not?

5. Classify each of the following energies as kinetic energy or potential energy:

(a) The energy in a chocolate bar.

(b) The energy of rushing water used to turn a turbine or a water wheel.

(d) The energy in a stretched rubber band.

6. Decide which of the following objects has more kinetic energy:

(a) A 200 lb. man running at 6 mph or a 200 lb. man running at 3 mph.

(b) A 200 lb. man running at 7 mph or a 150 lb. man running at 7 mph.

7. A car and a truck are traveling along the highway at the same speed.

(a) If the car weighs 1500 kg and the truck weighs 2500 kg, which has more kinetic energy, the car
or the truck?

(b) Both the car and the truck convert the potential energy stored in gasoline into the kinetic energy
of motion. Which do you think uses more gas to travel the same distance, the car or the truck?

8. You mix two chemicals in a beaker and notice that as the chemicals react, the beaker becomes
noticeably colder. Which chemicals have more chemical potential energy, those present at the start
of the reaction or those present at the end of the reaction?

Vocabulary

heat Energy that is transferred from one object to another object due to a difference in temperature.
Heat naturally flows from a hot object to a cooler object.

force Any push or pull.

work A force applied over a distance.

kinetic energy Energy associated with motion.

potential energy Stored energy. Potential energy depends on an object's position or mixture's compo-
sition.

chemical potential energy Potential energy stored in the atoms, molecules, and bonds of matter.

Law of Conservation of Energy Energy cannot be created or destroyed; it can only be changed from
one form to another.

Law of Conservation of Mass and Energy The total amount of mass and energy in the universe is
conserved.

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Image Sources

(1) US National Archives and Records Administration. A pony express rider, circa 1861.. Public
Domain.

(2
(3
(4
(5

(6
(7
(8
(9
(10

(11
(12

(13

(14

(15
(16
(17
(18
(19
(20
(21

CK-12, Richard Parsons. Shopping list chiseled on a rock.. CC-BY-SA.

US Dept. of Health and Human Services. Receiving a vaccination.. Public Domain.

A hot air balloon. CC-BY-SA.

CK-12 Foundation. With no large mass nearby, this spaceship is weightless.. CC-BY-SA.
http : //en . wikipedia . org/wiki/Image : Tehachapi_wind_f arm_2 . jpg. GNU-FDL.

A modern cell phone.. Public Domain.

Sharon Bewick. Some common molecules and the atoms that they are made from.. CC-BY-SA.
http : //www . f lickr . com/photos/theyoungthousands/1978086180/. CC-BY.

The Periodic Table.. CC-SA.

The type of balance a chemist might use to measure the mass of an object.. Public Domain.

Antoine Lavoisier, "The Father of Modern Chemistry.". Public Domain.

http : //en. wikipedia. org/wiki/Image : NGC_^l^._%28NASA-med%29 . jpg Everything from an ant
to an entire galaxy is composed of matter.. GNU-FDL, Public Domain.

Unknown. http://commons.wikimedia.Org/wiki/File:Wagon_train.jpg. Public Domain.

http : //en . wikipedia . org/wiki/Image : 3D_model_hydrogen_bonds_in_water .jpg. GNU-FDL.

Maggie Black's den medeltida kokboken. Physician letting blood from a patient.. Public Domain.

A photo of a wooden model of a Greek ship that has both sails and oars.. GNU-FDL.
Bryan. A modern jetliner.. CC-BY-SA.
http: //commons, wikimedia. org/wiki/File: Jiajing.jpg. Public Domain.

The first horse-drawn street car in Seattle, Washington in 1884-- Public Domain.

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Chapter 2

Chemistry - A Physical Science

2.1 Measurements in Chemistry

Lesson Objectives

Define qualitative and quantitative observations.

Distinguish between qualitative and quantitative observations.

Use Quantitative observations in measurements.

State the different measurement systems used in chemistry.

State the different prefixes used in the metric system.

Do unit conversions.

Use scientific notation and significant figures.

Use basic calculations and dimensional analysis.

Use mathematical equations in chemistry.

Introduction

One of the steps in the scientific method is observation. Observation involves recording data about the phe-
nomenon we wish to investigate. There are two different types of observations which are called qualitative
and quantitative. Qualitative observations are those involving words only while quantitative observations
are those involving both words and numbers. Although all the observations we can make on a phenomenon
are valuable, quantitative observations are more helpful than qualitative. Qualitative observations are
somewhat more vague in nature because they involve comparative terms.

For example, a qualitative observation would be "The attendance clerk is a small woman." If the observer
was 6 feet 4 inches tall, he/she might refer to a woman who is 5 feet 8 inches tall as "small". But if the
observer reported this observation to a person who was 5 feet 2 inches tall, the listener would not acquire
a good idea of the height of the attendance clerk because they would not think that a woman who is
5 feet 8 inches tall was small (Figure ??).

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My, that
woman is so
tall!

The description "a small woman" could refer to any woman whose height was between 2 feet and 6 feet
tall depending on who did the observing. Similarly, "a small car" could refer to anything from a compact
car to a child's toy car. The word small is a comparative term. The same is true of words like tall, fast,
slow, hot, and cold. These words do not have exact meanings.

Quantitative observations on the other hand, have numbers and units associated with them and are,
therefore, more exact (Table 2.1). Even if the number is only an estimate, it is more valuable than no
number at all.

Table 2.1: Qualitative and Quantitative Observations Comparisons

Qualitative (words only)

Quantitative (words and numbers)

The girl has very little money.
The man was short.
Use a small test tube.
It is a short walk to my house.

The girl has 85 cents.
The man was 5 feet2 inches tall.
Use a test tube than 12 cm long.
It is about 1 mile to my house.

You can see that even if the number is an estimate, a quantitative observation contains more information
because of the number associated with the observation. (Some people might not think that a walk of one
mile was short even though the speaker in the above case did. If an actual measuring instrument is not
available, the observer should always try to estimate a measurement so the observation will have a number
associated with it.

While estimated measurements may not be accurate, they are valuable because they establish an approx-
imate size for observations. The observation, "The car is small", provides us with certain information.
We know that the object is some kind of automobile (perhaps real, perhaps a toy) and we know that it is
probably smaller than a limousine because almost no one would describe a limousine as "small". Suppose
instead, the observation had been, "The car is about 3 feet tall, 3 feet long, and 2 feet wide." While these
estimated measurements cannot be considered to be accurate, we now know that we are not dealing with
a compact automobile nor are we dealing with a toy car like Hot Wheels. With these estimated measure-

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merits, we know we are dealing with a car that is about the size of a tricycle. If we discover later that
the car was actually 2 feet high instead of 3 feet high, it is not a problem because we knew the original
observation was an estimate since it contained the word "about". Estimates are excellent observations if
we do not have the ability to actually measure the object. Estimated measurements qualify as quantitative
observations.

Here is some information you may find helpful in making estimated observations. The distance from the
top of your index finger to the first knuckle is about one inch. The entire index finger is about three inches
long. Your foot is probably between eight and twelve inches long. Your height is probably between five
and six feet. The distance from your wrist to your elbow is about one foot. A twenty-five cent coin is
about one inch across and dollar bills are about 2.5 inches wide and about six inches long (Figure 2.1).

Figure 2.1: Helpful hints when making estimates.

In science, accurate quantitative observations are a great deal more useful than qualitative observations.
When we speak about accurate quantitative measurements, we are essentially referring to measurements.
Accurate measurements are vital to science. There are many measurement systems used in the world but
only one that is used consistently in science. That system is called the International System of Units and
is abbreviated as SI units. You are probably already familiar with part of the SI system because part of
the SI system is called the Metric System.

Mass and Its SI Unit

When you step on a bathroom scale, you are most likely thinking that about determining your weight,
right? You probably aren't wondering if you have gained mass. Is it okay then to use either term?

Although we often use mass and weight interchangeably, each one has a specific definition and usage. The
mass of an object does not change; whether the object is on the earth's equator, on top of Mt. Everest,
or in outer space, the mass will always be the same. Because mass measures how much matter the object
contains, it has to be a constant value.

Weight, on the other hand, is a measure of the force with which an object is attracted to the earth or body
upon which it is situated. Since the force of gravity is not the same at every point on the earth's surface,
the weight of an object is not constant. For example, an object weighing 1.00000 lb in Panama weighs
1.00412 lb in Iceland. For large objects this difference may not be significant. However, since we will often
be working with extremely tiny pieces of matter - atoms, molecules, etc. - we need to use mass and not
weight.

The basic unit of mass in the International System of Units (SI comes from the French name, Systeme
Internationale) is the gram. A gram is a relatively small measurement compared to, for instance, one
pound. 454 grams equals one pound. While pounds are helpful in measuring the mass of a package that
needs to be mailed, grams are much more useful in science.

One gram is equal to 1,000 milligrams or 0.001 kilogram; there are numerous intermediate measurements
between each of these mass units as well as ones that are even larger and smaller that may be appropriate
to the application at hand. These will be discussed in more detail in a later section.

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Length and its SI Unit

When the four minute mile was achieved on May 6, 1954 by Roger Bannister, it was an international
sensation. Today, many runners have broken that record. Only a few countries measure length or distance
using miles, feet or inches. For instance, if you live in the US, you probably know your height in feet and
inches, right? Or, if there is a mountain or even a hill near where you live, you probably know its height in
feet. And when you discuss how far school is from your home, you probably try to figure out the distance
in miles.

However, most of the world measures distances in meters and kilometers; for shorter lengths, millimeters
and centimeters will be used. For a student in Germany, she will state how many kilometers her school
is from home, and the height of the mountain she is thinking of climbing will be given in meters (Figure
2.2).

Figure 2.2: Using a laser to make a precise distance measurement.

Because the metric system is a decimal system, changing between the various measurements simply becomes
a matter of moving a decimal point to the right or left. We will discuss this in greater depth later on.

While a gram is a relatively small measurement, a meter is quite similar in length to a measurement with
which we are familiar, a yard. Specifically 1 meter is equal to 1.1 yards or 39.4 inches ; one kilometer is
equal to 0.621 miles; and one inch is equal to 2.54 centimeters.

Volume: A Derived Unit

Volume is used to measure how much space an object takes up. It is a derived unit, meaning it is based on
another basic SI unit- in this case, the meter (length) was used to measure the sides of a cube, designating
a certain volume. This volume was determined to be a cubic meter, m 3 , which is used as the standard SI
unit of volume. This is a very large unit, and it is not very useful for most measurements in chemistry.
A more common unit is the liter (L), which is equal to 1/1000 of a cubic meter. Another commonly used
volume measurement is the milliliter; 1000 L = 1 L.

One liter is the volume of the soda bottle that you might have recently purchased and have sitting in the
refrigerator at home. You might also have a quart of milk in your refrigerator. Even though the size of the
liter container and the milk carton may not appear to be the same, they are, in fact, almost exactly the

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same volume. A quart is just slightly smaller in volume than a liter (1 L = 1.057 quarts). It's only the
packaging that is different!

Measuring Temperature

In order to discuss temperature scales, let's briefly compare the concepts of heat and temperature. Heat
is a measurement of the total amount of kinetic energy while temperature describes the intensity of the
heat, or what is often referred to as the average kinetic energy of the material. When we are measuring
the temperature of an object we are measuring its average kinetic energy. For that, we use the Celsius and
kelvin scales. Scientists do not usually use the Fahrenheit scale. The size of a degree in kelvin is the same
as 1 degree Celsius. The difference is that the kelvin scale begins with an absolute zero, the temperature
at which all motion stops. To convert between the two scales you can use: K =° C + 273. Therefore, on
the kelvin scale, water freezes at 273 K and boils at 373 K.

You might want to make note of the following: while most mathematical calculations in chemistry require
you to convert Celsius temperatures into kelvin, when you are given a difference in temperature, AT , you
do not need to convert it to kelvin! A difference in temperature is the same whether it is in Celsius or
kelvin.

Lesson Summary

• The International System unit for mass if the gram.

• The International System unit for length is the meter.

• The unit for volume is derived from a cube that is 1.00 meter on each side; therefore the volume unit
is cubic meters. A more common unit is the liter which is j^ of a cubic meter.

• The SI uses both °C and absolute temperature in Kelvin (K) for temperature units. K =° C + 273.

Review Questions

1. What are the basic units of measurement in the metric system for length and mass?

2. What unit is used to measure volume? How is this unit related to the measurement of length?

3. Explain the difference between weight and mass.

4. Give both the Celsius and Kelvin values for the boiling and freezing points of water.

5. How do you convert from Celsius to Kelvin? How does one degree Celsius compare with one Kelvin?

6. If someone told you that a swimming pool's temperature was 275 K, would it be safe for you to go
for a swim?

7. Determine which metric measurement you would use for each of the following:

(a) The distance to the moon.

(b) The mass of a donut.

(d) The length of your little finger.

Further Reading / Supplemental Links

• Website with lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 1-3 is
on Measuring Matter, http://www.fordhamprep.org/gcurran/sho/sho/lessons/lessonl3.htm

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Vocabulary

International System of Units, SI The SI system of units is the modern form of the metric system
and is generally a system devised around the convenience of multiples of 10.

Kelvin temperature scale The kelvin is unit of increment of temperature and is one of the seven SI
basic units. The Kelvin scale is thermodynamic absolute temperature scale where absolute zero is
the absence of all thermal zero. At K = 0, there is no molecular motion. The kelvin is not referred
to as a "degree", it is written simply as K, not ° K.

2.2 Using Measurements

Lesson Objectives

• Understand the metric system and its units.

• Convert between units.

• Use scientific notation in writing measurements and in calculations.

• Use significant figures in measurements.

Introduction

The metric system is a decimal system. This means that making conversions between different units of
the metric system are always done with factors of ten. Let's consider the English system - that is, the
one that is in everyday use in the US as well as England - to explain why the metric system is so much
easier to manipulate. For instance, if you need to know how many inches are in a foot, you only need to
remember what you at one time memorized: 12 inches = 1 foot. But now you need to know how many
feet are in a mile. What happens if you never memorized this fact? Of course you can look it up online or
elsewhere, but the point is that this fact must be given to you as there is no way for you to derive it out
yourself. This is true about all parts of the English system: you have to memorize all the facts that are
needed for different measurements.

The Metric System

In the metric system, you need to know (or yes, memorize) one set of prefixes and then apply them to each
type of measurement. Then if a larger measurement is needed, such as kilometers, but you have used a
meter stick, you only need to move the decimal to convert the units.

Example: If you have measured the distance as 60.7 meters, what is the length in kilometers?

Solution: 60.7 meters = 0.0607 kilometers since there are 1,000 meters in 1 kilometer.

Not only can you easily convert from kilometers to meters, but conversions, such as liters to cubic meters,
are also easy. Try converting from cubic feet to gallons! All metric system conversions simply require the
moving of the decimal and/or adding zeros. You don't even need a calculator. On the other hand, if you
had to convert from miles to inches, not only would you have to remember all of the conversion factors,
but you would probably also need a calculator to make the conversion.

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Metric Prefixes

The metric system uses a number of prefixes along with the base unit. To review: the basic unit of mass
is a gram (g), that of length is meter (m), and that of volume is liter (L). When the prefix centi is place in
front of gram, as in centigram, the measure is now y^ of a gram. When milli is placed in front of meter,
as in millimeter, the measure is now y^ of a meter. Common prefixes are in Table 2.2:

Table 2.2:

Prefix Meaning Symbol

pico- 1CT 12 p

nano- 10~ 9 n

micro- 10 -6 fi

milli- 10~ 3 m

centi- 10~ 2 c

deci- 1CT 1 d

kilo- 10 3 k

Unit Conversions

Making conversions in the metric system is relatively easy: you just need to remember that everything is
based on factors of ten. For example, let's say you want to convert 0.0856 meters into millimeters. Looking
at the chart above, you can see that 1 millimeter is 10 -3 meters; another way to say this is that there are
1000 millimeters in one meter. You can set up a mathematical expression as follows

(0.0856 m) X (1000 mm/1 m) =? mm

When you solve this equation, you first want to see which units to divide out. In this case, you notice
meters appear in both the numerator and denominator, so you will be able to cancel them.

(0.0856 m) X (1000 mm/1 m) =? mm

Now all that is left to do is multiply 0.0856 by 1000. To do this, you are just going to move the decimal
point three places to the right:

(0.0856 m) X (1000 mm/1 m) = 85.6 mm

Example: Convert 153 grams to centigrams.

Solution: 153 grams X cen lgram& — 15300 centigrams

Scientific Notation

Scientific notation is a way to write very large or very small numbers (Figure 2.3); the decimal point is
moved so that there is one digit in the unit's position and all of the decimal places are held as a power
of ten. This is important in chemistry because many of the measurements we make either involve very
large numbers of atoms/molecules or very tiny measurements, such as masses of electrons or protons. For
example, consider a number such as 839,000,000. While this number is not too difficult to write out, it
is more conveniently written in scientific notation. Written in scientific notation, this number becomes:
8.39 x 10 8 . The "10 8 " means that ten is multiplied by itself eight times: 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10.
As you can see, writing 10 8 is much easier!

We can also use scientific notation to write very small numbers. Take a number such as 0.00000481. It is

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Figure 2.3: Even using large distance units such as kilometers, you would still need to use scientific notation
to measure the size of this galaxy.

easy to make mistakes in counting the number of zeros in this number. Also, many calculators only let you
enter in a certain number of digits. When we write this in scientific notation, it is important to notice that
the measurement is less than one, therefore, the exponent on 10 will be negative: this number becomes
4.81 x 10~ 6 . In this case, the decimal point was moved six places to the right.

It is important that you know how to perform calculations using numbers written in scientific notation.
For example, the following problem shows two numbers with exponents being multiplied together:

(2.90xl0 3 )(1.60xl0 6 ) =?

To solve this problem, you would multiply the terms (2.90 and 1.60) like you normally would; then you
would add the exponents:

2.90x1.60 = 4.64
10 3 x 10 6 = 10 9

Therefore, combining these values gives the answer 4.64 x 10 9 .

Significant Figures

The tool that you use determines the number of digits that will be in a measurement. For example, if you
say an object has a mass of "5 kg", that is not the same as saying it has a mass of "5.00 kg" since you
must have measured the masses with two different tools - the two zeros in "5.00 kg" would not be written
if the tool that was used could not measure to two decimal places. Even though the mass seems to be the
same, the uncertainty of the measurement is not. When you say "5 kg," that means you have measured
the mass to within +/ - 1 kg. The actual mass could be 4 or 6 kg. For the 5.00 kg measurement, you have
measured the mass to within +/ - 0.01 kg, so the actual mass is between 4.99 and 5.01 kg.

Using Significant Figures in Measurements

How do you know how many significant figures are in a measurement? General guidelines are as follows:
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• Any nonzero digit is significant [4.33 has three significant figures].

• A zero that is between two nonzero digits is significant [4.03 has three significant figures].

• All zeros to the left of the first nonzero digit are not significant [0.00433 has three significant figures] .

• Zeros that occur after the decimal are significant. [40.0 has three significant figures. The zero after
the decimal point tells us that the value was measured to the tenths place] .

• Zeros that occur without a decimal are not significant [4000 has one significant figure since the zeros
are holding the 4 in the thousands position].

Examples:

1) How many significant figures are in the number 1.680?

Solution: There are three nonzero digits and one zero appears after the decimal point. Therefore, there
are four significant figures.

2) How many significant figures are in the number 0.0058201?

Solution: There are 4 nonzero digits and 1 zero between two numbers. Therefore, there are 5 significant
figures. The first three zeros are not significant since they are simply holding the number away from the
decimal point.

Lesson Summary

• The metric system is a decimal system; all magnitude differences in units are multiples of 10.

• Unit conversions involve creating a conversion factor.

• Very large and very small numbers are expressed in exponential notation.

• Significant figures are used to express uncertainty in measurements.

Review Questions

1. Convert the following linear measurements:

(a) 0.01866 m = cm

(b) 2156 mm = m

(d) 1250.2 m = km

2. Convert the following mass measurements:

(a) 155.13 mg — kg

(b) 0.233 g = mg

(d) 0.2885 g = mg

3. Write the following numbers in scientific notation:

(a) 0.0000479

(b) 251,000,000

(d) 0.00206

4. How many significant figures are in the following numbers?

(a) 0.006258

(b) 1.00

(d) 12500

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Further Reading / Supplemental Links

The website listed below offers lessons, worksheets, and quizzes on many topics in high school chemistry.

• Lesson 1-7 is on Temperature Conversion.

• Lesson 2-1 is on the International System of Measurements.

• Lesson 2-3 is on Significant Figures.

• Lesson 2-5 is on Scientific Notation, http://www.fordhamprep.org/gcurran/sho/sho/lessons/
lessonl2.htm

Vocabulary

scientific notation A shorthand way of writing very large or very small numbers. The notation consists
of a decimal number between 1 and 10 multiplied by an integral power of 10. It is also known as
exponential notation.

significant figures Any digit of a number that is known with certainty plus one uncertain digit. Begin-
ning zeros and placeholder zeros are not significant figures.

2.3 Using Mathematics in Chemistry

Lesson Objectives

• Use units in problem solving.

• Do problem solving using dimensional analysis.

• Use significant figures in calculations.

Introduction

Unit terms are the words following a measurement that tell you on which standard the measurement
is based. Every measurement must have a unit term. The unit terms also follow the algebraic rules
of exponents and cancellation. Carrying the unit terms through mathematical operations provide an
indication as to whether the mathematical operation was carried out correctly. If the unit term of the
answer is not correct, it is an indication that the mathematical operation was not done correctly.

Using Units in Problem Solving

Anytime we have to do a calculation, it is important to include the units along with the actual numbers.
One reason is that you can often decide how to solve the problem simply by looking at the units. For
example, let's say you are trying to calculate solubility. One of the units used for solubility is grams/liter
(g/L). Imagine that you have forgotten how to do the mathematical calculation for this problem, but you
have measured how many grams of a solid dissolved into a certain number of liters. Looking at the units
of the values that you have (g and L) and at the units of the answer you want to get (g/L), you can figure
out the mathematical set-up. The g/L unit allows you to know it needs to be "grams divided by liters."

You will also note that as you do a calculation, you will be working with units in a similar manner as you
would a number. Just as with numbers, units can be divided out when that specific unit appears in the
numerator as well as the denominator.

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As a final note on units, think of them in an "apples and oranges" context. You can't subtract meters from
kilometers without first converting the measurements into common units. Always check a measurement's
units to make sure that they are appropriate for a given calculation.

Using Conversion Factors

Conversion factors are used to convert one unit of measurement into another. A simple conversion factor
can be used to convert meters into centimeters, or a more complex one can be used to convert miles per
hour into meters per second. Since most calculations require measurements to be in certain units, you will
find many uses for conversion factors. What always must be remembered is that a conversion factor has to
represent a fact; this fact can either be simple or much more complex. For instance, you already know that
12 eggs equal 1 dozen. A more complex fact is that the speed of light is 1.86 x 10 5 miles/sec. Either one
of these can be used as a conversion factor depending on what type of calculation you might be working
with. The following section provides you with more examples of this.

Dimensional Analysis

When using conversion factors (and for that matter, a lot of other calculations), a process called dimensional
analysis is extremely useful. Dimensional analysis allows you to make a number of unit conversions in a
single calculation. It will also help you keep the units straight.

Example: A car travels 58.5 miles, using 1.5 gallons of gasoline. How do you express this in kilometers/liter?
You know that there are 3.78 liters in a gallon, and a kilometer is 0.62 miles. How would you make this
conversion?

Solution:

You first need to write out a mathematical expression showing all your conversion factors and units:

58.5 miles
1.5 gallons

)/ 1 gallon \ /l kilometer \
X \3.78 liters/ X \ 0.62 miles J

Next, you need to check for units to divide out:

58.5 miles

1.5 gallons

)/ 1 gallon \ /l kilometer \
X \3.78 liters/ X \ 0.62 miles /

Notice that at this point you are left with kilometers in the numerator and liters in the denominator. Your
last step is to multiply your numbers, and your answer will be in kilometers/liter:

'58. 5\ / 1 \ /l kilometer\ /16. 64106503 kilometers\
x

(ttH

3.78 liters/ \ 0.62 / \ liter /

This is the answer that your calculator will give to you. However it is not the correct answer. For that we
have to proceed to the next section:

Using Significant Figures in Multiplication and Division

Whenever we do a calculation, we need to pay attention to the significant figures. The rule is that your
final answer can only be as precise as your least precise measurement. This means that the least precise
tool used for any measurement in the calculation will determine how precise the answer will be.

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For multiplication and division, first determine the number of significant figures in each of the measure-
ments; the number of significant figures in your answer will be the same as the least number in the
calculation. For example, if you multiplied the number 1.02584 by 2.1, your answer can only have two
significant figures. The same rule applies for division.

Example: Divide the number 125.688 by 14.01. Express your answer using correct significant figures.

Solution: 125.688 has 6 significant figures, and 14.01 has 4. Therefore, your answer can only have 4
significant figures.

14.01

An important point to remember is that if you are multiplying or dividing by an exact number, then you
treat that number as having an infinite number of significant figures. An exact number is a number that
is written without all of its known significant figures. For instance, one meter or one dozen have many
significant figures but we just don't write all of them, so these kind of measurements never determine the
number of significant figures in a calculation. There is another type of exact number and that is using
a measurement such as five people, ten dogs, or one cat. Again, these do not determine the number of
significant figures in an answer.

Example: You have one dozen identical objects, with a total mass of 46.011 grams. What is their average
mass?

Solution:

12 = 3.83425, which rounds off to 3.8342 (5 significant figures, the same as 46.011)

Using Significant Figures in Addition and Subtraction

There is a different rule for determining significant figures when adding or subtracting measurements. Now,
you will need to look for the measurement with the least number of significant figures to the right of the
decimal place; this number of decimal places will determine the number of significant figures to be used in
the answer.

Example: What is the sum of 14.3 and 12.887?

Solution:

14.3 + 12.887 = 27.187
The number 14.3 only has 1 digit to the right of the decimal point, so our answer is rounded off to 27.2.

Lesson Summary

• Dimensional analysis aids in problem solving.

• Conversion factors are created by unit analysis.

• Significant figures must be carried through mathematical operations.

• The answer for an addition or subtraction problem must have digits no further to the right than the
shortest addend.

• The answer for a multiplication or division problem must have the same number of significant figures
as the factor with the fewest significant figures.

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Review Questions

1. Perform the following calculations and give your answer with the correct number of significant figures:

(a) 0.1886x12

(\y) 2.995
\ u ) 0.16685

VW 0.1223

(d) 910x0.18945

2. Perform the following calculations and give your answer with the correct number of significant figures:

(a) 10.5 + 11.62

(b) 0.01223 + 1.01

3. Do the following calculations without a calculator:

(a) (2.0xl0 3 )(3.0xl0 4 )

(b) (5.0xl0- 5 )(5.0xl0 8 )

,,n (3.0xlQ- 4 )(2.0xlQ- 4 )
W 2.0X10- 6

Further Reading / Supplemental Links

The website listed below offers lessons, worksheets, and quizzes on many topics in high school chemistry.

• Lesson 2-4 is on the Factor-Label Method of Unit Conversion, http://www.fordhamprep.org/
gcurran/sho/sho/lessons/lessonl2 . htm

Vocabulary

dimensional analysis A technique that involves the study of the dimensions (units) of physical quanti-
ties. It affords a convenient means of checking mathematical equations.

2.4 Using Algebra in Chemistry

Lesson Objectives

• Be able to rearrange mathematical formulas for a specific variable.

• Have an understanding of how to use units in formulas.

• Be able to express answers in significant figures and with units.

Introduction

During your studies of chemistry (and physics also), you will note that mathematical equations are used
in a number of different applications. Many of these equations have a number of different variables with
which you will need to work. You should also note that these equations will often require you to use
measurements with their units. Algebra skills become very important here!

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Solving Formulas with Algebra

Sometimes, you will have to rearrange an equation to get it in the form that you need. When you are
working with an equation such as D = y (density = mass/volume) and asked to solve for density, it is
relatively easy; all you have to do is substitute the measurements and solve - of course, keeping in mind
significant figures!

If you are asked to solve the above equation for M, then you will need to manipulate the equation to isolate
the desired variable, in this case in the form of "M =." To do this, you will need to move the V from the
right side to the left side of the equation. As the V is in the denominator, you will need to multiply both
sides of the equation by V:

VXD=(V)(£)

Multiplying this out:

VxD = M

A similar process is used if you need to solve for V:

Take the previous equation (V x D = M) and divide both sides by D:

Solving, this becomes:

What if you are given a more complex equation, like PV = nRT, and asked to solve for "«"? You need to
follow the same steps as you did in the above examples. The only difference is that there are more symbols
to rearrange.

Solution:

Look at the original equation: PV — nRT . Our goal is to get "n" on one side of the equation by itself.

To remove the RT from the right side, we will divide both sides by RT:

PV nRT

RT RT

After solving, we are left with:

n —

Algebra with Units and Significant Figures

So far, you've learned about units, significant figures, and algebraic manipulation of equations. Now it's
time to put all three of these together. We'll start with a simple example: density. Density is a measure
of the amount of mass per unit of volume, and a common unit used is g/niL. In the first example, we're
going to do a straightforward calculation of density from a given mass and volume.

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Example 1

What is the density of an object that has a mass of 13.5 g and a volume of 7.2 mL?

Solution:

The equation is as follows:

mass
D

volume

Substituting in the known values (with units): D = 72m L

Finally, solving the equation and rounding off the answer based in significant figures:

13 5 e

D = = 1.9g/mL(2 significant figures)

7.2 mL b/ v s s i

This calculation was easy to do because there was no rearranging of the equation and no cancellation of
units.

For the next example, let's look at a more complex calculation.

Example 2:

A sample of an ideal gas has a volume of 14.2 L at a pressure of 1.2 atm. If the gas pressure is increased
to 1.8 atm, what is the new volume?

Solution:

This problem uses Boyle's law:

Pi Vi = P 2 V 2

All the variables are known except for V 2 , so the equation needs to be rearranged to solve for the one
unknown. We can do this by dividing both sides by P 2 :

P1V1

Now we substitute in the known values and their units:

(1.2 atm) (14.2 L)

(1.8 atm)

Next, we cancel out units:

(1.2)(14.2 L)

Finally, we calculate our answer and round off to the appropriate number of significant figures:

V 2 = 9.5 L

Lesson Summary

• Students of chemistry need to be able to use algebra in their calculations.

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V 2

Review Questions

1. For the equation PV = nRT, re- write it so that it is in the form of " T =."

2. The equation for density is D = M/V. If D is 12.8 g/cm , and M is 46.1 g, solve for V, keeping
significant figures in mind.

3. The equation P\ V\ = Pi V2, known as Boyle's law, shows that gas pressure is inversely proportional
to its volume. Re-write Boyle's law so it is in the form of V\ =?.

4. The density of a certain solid is measured and found to be 12.68 g/mL. Convert this measurement
into kg/L.

5. In a nuclear chemistry experiment, an alpha particle is found to have a velocity of 14,285 m/s.
Convert this measurement into miles/hour.

Further Reading / Supplemental Links

• College Chemistry- Schaum's Outlines. Jerome Rosenberg and Lawrence Epstein, McGraw-Hill,
1997.

• Chemistry, 7 th Edition, Raymond Chang, McGraw-Hill, 2002.

• Chemistry, J. Dudley Herron, Jerry L. Sarquis, Clifford L. Schrader, David V. Frank, Mickey Sarquis,
David A. Kukla. D.C. Heath and Co., 1993.

• World of Chemistry, Steven S. Zumdahl, Susan L. Zumdahl, and Donald J. DeCoste. McDougal
Littell, 2007.

Labs and Demonstrations for Using Algebra in Chemistry
Density Determination

Pre-Lab Discussion

Density is defined as the mass per unit volume of a substance. The table below lists the densities of some
well known substances.

Table 2.3: Density of Some Common Substances
Substance Density Substance Density

Water

1.0 g/cm 3

Aluminum

2.7 g/cm 3

Oxygen gas

0.0013 g/cm 3

Iron

8.9 g/cm 3

Sugar

1.6 g/cm 3

Lead

11.3 g/cm 3

Table salt

2.2 g/cm 3

Gold

19.3 g/cm 3

Density measurements allow scientists to compare the masses of equal volumes of substances. If you had
a piece of lead as large as your fist and a piece of gold as large as your thumb, you would not know which
substance was innately heavier because the size of the pieces are different. Determining the density of the
substances would allow you to compare the masses of the same volume of each substance. The process for
finding the density of a substance involves measuring the volume and the mass of a sample of the substance
and then calculating density using the following formula.

mass in grams

Density — —

volume in mL

Example: Calculate the density of a piece of lead whose mass is 226 grams and whose volume is 20.0 mL.
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Also calculate the density of a sample of gold whose mass is 57.9 grams and whose volume is 3.00 mL.
Solution:

Density of Lead
Density of Gold

mass 226 g

volume 20.0 mL

mass 57.9 g

volume 3.00 mL

11.3 g/mL
19.3 g/mL

Methods of Measuring Mass and Volume

The mass of substances is measured with a balance. In the case of a solid object that will not react with
the balance pan, the object may be placed directly on the balance pan. In the case of liquids or reactive
solids, the substance must be placed in a container and the container placed on the balance pan. In order
to determine the mass of the substance, the mass of the container is determined before hand (empty) and
then the container's mass is subtracted from the total mass to determine the mass of the substance in the
container. There are several common procedures for determining the volume of a substance. The volume
of a liquid is determined by pouring the liquid into a graduated cylinder and reading the bottom of the
meniscus. For a regularly shaped solid, the volume can be calculated from various linear measurements.

CUBE

SPHERE

CYLINDER

V=(x)(y)(z)

V=lnr 3

V=nr 2 h

For an irregularly shaped object, a graduated cylinder is partially filled with water and the volume mea-
sured. The object is then submerged in the water and the new volume measured. The difference between
the two volumes is the volume of the submerged object.

Equipment: Specific gravity blocks, graduated cylinders (10 mL and 100 mL), thread, millimeter ruler,
balance, distilled water, glycerol. (If you have an overflow can, it also works well for submersion.)

Procedure:

1. Obtain a regularly shaped object from your teacher. Measure and record its mass in grams and its
dimensions in centimeters.

2. Add approximately 50 mL of tap water to a 100 - mL graduate and record its exact volume. Tie a
thread to the block and carefully immerse it in the cylinder of water. Record the new volume in the
cylinder.

3. Measure and record the mass of a clean, dry, 10 - mL graduated cylinder.

4. Add exactly 10.0 mL of distilled water to the cylinder. Measure and record the combined mass of the
cylinder and water.

5. Repeat steps 3 and 4 with glycerol instead of water.

Data Table

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Object

code name or letters

width

height

length

volume of water before block
volume of water after block _
Distilled Water

volume

mass of empty graduate

combined mass

Glycerol
volume

mass of empty graduate

combined mass

Calculations

1. Find the volume of the solid object using the dimensions and appropriate formula.

2. Find the volume of the solid object using water displacement.

3. Find the density of the block using the volume from calculation 1.

4. Find the density of the block using the volume from calculation 2.

5. Find the density of the distilled water.

6. Find the density of the glycerol.

7. If your teacher gives you the accepted values for the densities in this lab, calculate the percent error
for your values.

(experimental value) — (actual value)

% error = — X 100

[actual value)

Image Sources

(1) Using a laser to make a precise distance measurement.. CC-BY-SA.

(2) http://www.flickr.com/photos/pingnews/474783096/. Public Domain.

(3) Richard Parsons. Helpful hints when making estimates.. CC-BY-SA.

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Chapter 3

Chemistry in the Laboratory

3.1 Making Observations

Lesson Objectives

• Define qualitative and quantitative observations.

• Distinguish between qualitative and quantitative observations.

• Use Quantitative observations in measurements.

Introduction

Take out a piece of paper and record a chart similar to the one below. Look up from this text and scan
the room. Write down what you see around you in as much detail as you feel necessary to remember it
after you walk away. Better still, you want to be able to show your chart to someone who can then be able
to picture where in the room you were sitting. A chart is usually necessary to record these observations
simply for organization (Table 3.1).

Table 3.1:

Item Observation

2.
3.

4.
5.

Science depends on keeping records of observations for later interpretations. These interpretations may
lead to the development of theories or laws. Without accurate observations, scientists cannot make any
interpretations and therefore cannot draw conclusions.

Here is a simple test for you. Pretend we're visiting a forensic scientist, hired to investigate the scene of a
crime. You are only asked to analyze the observations gathered by the other scientists at the scene. You
must try not to make too many assumptions yet try and make your decision based on the data at hand.
One summer evening, Scott and Brenda came home from work to find their house in shambles. Neighbors,

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friends, and colleagues are baffled by the strange occurrence. The television was found on in the house.
Food was on the table ready to be eaten. All of Scott's coin collections, his precious metals, and Brenda's
prized possession - her statue of Galileo - were gone. Foul play is suspected. The lead investigator gives
you the following observations gathered from the scene and suspects (Table 3.2):

Table 3.2:

Observations at Scene Suspect 1: 180 lb male Suspect 2: 220 lb male Suspect 3: 120 lb female

Blood type = B
Fiber sample =
polyester

Powder found = white
Shoe Print found =
work boot

Blood type = B
Sweater = polyester

Blood type = B
Blazer = wool knit

Works in sugar factory Pastry chef

Would not comply
Pants = polyester

Car sales woman

From the table, can you deduce who might have been involved in the alleged crime? Do you need more
information? How good are the observations in order for you, the scientist, to make accurate conclusions?
What will you base your decision on? What other information do you need? Remember someone will be
charged for this crime. Observations are the key to science all around us and in our everyday lives.

Qualitative Observations

Science is full of observations but of two different types. What we see, smell, feel, and hear are observations
that scientists depend on to determine whether chemical reactions have been occurring or have come to
completion. This is one type of observation known as a qualitative observation. Qualitative observations
give the descriptive properties of a substance or being and therefore are without numbers.

When you made your table above, what kind of observations did you make? Take a look at the table. Did
you note any colors from the surroundings? Was there a window nearby? If so, was it open? Did you
happen to hear any sounds from outside the window? Did you see a vehicle drive by? If so, what color
was it? Take a look at the sample question below and see if you can determine the qualitative observations
from each of the figures in the question.

Sample Problem: List the qualitative observations for each of the figures below.

Left image: Fog caused by dry ice. (Source: http://commons.wikimedia.Org/wiki/File:DryIceSublimation.
JPG. Public Domain)

Center image: Tulips. (Source: http://c0mm0ns.wikimedia.0rg/wiki/File:Vi0lett_tulips.jpg. Pub-
lic Domain)

Right image: Soda. (Source:http : //en. wikipedia. org/wiki/Image: Soda_bubbles_macro.jpg. Public
Domain)

Solution:

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Fog caused by dry ice: fog or smoke coming from top of the cup pouring over onto the tabletop

Tulips: purple tulips, sky with clouds in the background

Soda: bubbles, almost looks like effervescence from soda pop when you pour a glass of cola

Quantitative Observations

Sometimes qualitative measurements are enough to give an accurate representation of the events occurring.
In other cases, scientists need more information than what the senses offer in order to make correct
interpretations and then conclusions. Say, for example, there was a window in your classroom and outside
the window you see a battleship gray car zoom by, speeding you suspect because behind him you hear the
sirens of a police car. You make the interpretation that the car is speeding because of the police sirens.
The police could, in fact, be on another call. The only real accurate qualitative observations you can make
here are that you see a battleship gray car drive by, you hear the sirens of the police car, and you see the
police car drive by with sirens on. Now if you say, happen to have a hand held radar gun, and could then
measure that the car was travelling 50 mph in a 35 mph zone, then you could conclude the police were
chasing the car.

Quantities are a useful strand of observations. When you have observations that involve the use of numbers,
we refer to these as quantitative observations because they have amounts. In our car chase example above,
the measurement of 50 mph and 35 mph are both quantitative measurements. If we said that it is 85° F
outside in British Columbia, the temperature of 85° F is a quantitative observation. Now, if we said that
it is 85° F outside in sunny British Columbia, the temperature of 85° F is a quantitative observation, and
the "sunny" is a qualitative observation. See how it works? Now you try.

Sample Question:

Pick out the quantitative and qualitative observations from each phrase:

1. 3 g of NaCl dissolves in 10 mL of H^O to make a clear solution.

2. The spider on the wall only has seven legs remaining but is still big and hairy.

3. When 0.5 mL of a solution is put into a flame, the flame turns a brilliant green.

Solution:

1. Quantitative

2. Quantitative

3. Quantitative

3 g 10 mL; qualitative: clear,
seven; qualitative: big, hairy.
0.5 mL; qualitative: brilliant green.

Lesson Summary

• Qualitative observations describe the qualities of a substance or event.

• Qualitative observations are made using the senses (except taste - used only when appropriate).

• Qualitative observations do not involve numbers.

• Quantitative observations describe the quantities of a substance or event.

• Quantitative observations use numbers in the descriptions for the substance or event.

• Observations, either qualitative or quantitative, or both, are used as tools by scientists to make
representations and then interpretations about the surroundings.

Review Questions

1. Indicate in the following chart (Table 3.3) whether the observation is qualitative or quantitative.

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Table 3.3:

Number

Observation

Qualitative or Quantitative

4.
5.
6.

7.
8.

10.
11.

12.

The temperature of this room is

25°C.

It is comfortably warm in this

room.

Most people have removed their

coats.

The building is 25 stories high.

It is a very tall building.

The building is taller than any

nearby trees.

The bottle is green.

The bottle contains 250 mL of

liquid.

Robert bought his son a small

car.

The car is smaller than his hand.

The car is about three inches

long.

The race is about 27 miles long.

Vocabulary

qualitative observations Describe the qualities of something and are described without numbers.
quantitative observations Observations that involve the use of numbers (quantities).

Further Reading / Supplemental Links

• http://en.wikipedia.org/wiki

3.2 Making Measurements

Lesson Objectives

• Match equipment type based on the units of measurements desired.

• Determine significant figures of the equipment pieces chosen.

• Define accuracy and precision.

• Distinguish between accuracy and precision.

Introduction

As we learned in the previous section, qualitative observations require the use of the senses to gather data
in order to interpret what is happening in our surroundings and then make conclusions based on these

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interpretations. Quantitative observations gather data by using measurements. From these measurements
we can interpret the data and draw conclusions. How exactly do scientists gather all of this numerical
data? What kind of equipment is necessary and for what purposes? How accurate is it? Let's take a
look, first at some of the typical equipment used in chemistry and then at the skills necessary to determine
accuracy and precision. Let's explore the quantitative side of chemistry.

Equipment Determines the Unit of the Measurement

Think of the last laboratory experiment that you did. What kind of equipment did you use? If you were
measuring out a volume of a liquid, did you use a beaker or a graduated cylinder?

Beakers used in experiments. (Source: http://c0mm0ns.wikimedia.0rg/wiki/File:Beakers.jpg. CC-
BY-SA)

100 mL graduated cylinder. (Source: http://en.wikipedia.Org/wiki/Image:Graduated_cylinder.
jpg. Public Domain)

Look at the two figures above; if you were required to measure out 65 mL, what instrument would you most
likely want to use? The graduated cylinder has graduations every 10 mL and then further graduations
every 5 mL. The beakers could have graduations every 10 mL, 50 mL, or 100 mL depending on which
type you use. It would be easier to measure out the volume in a graduated cylinder. What if, in this same
lab, you needed to mass out 3.25 g of sodium chloride, NaCl. Look at the two figures below and determine
which piece of equipment you would use.

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A pan balance. (Source: http : //en . wikipedia . org/wiki/File : Balance_ / C3°/„AO_tabac_1850 . JPG . GNU-
FDL)

A digital balance. (Source: http : //commons . wikimedia . org/wiki/File : Digi-keukenweegschaall284 .
JPG. GNU-FDL)

'*dkt

p'tuuf firiVFfl

'UniillPMLVUl

■^

^r~

^^^^^^^^^^— ^

' aarw; *

^^"^

The pan balance measures only to +/ - 0.1 g. Therefore, you would have to mass out 3.3 g of NaCl rather
than 3.25 g. The digital balance can measure to +/ - 0.01 g. With this instrument you could measure
exactly what you need, depending on your skill of course!

The equipment you choose also determines the units in your measurement and vice versa. For example,
if you are given graduated cylinders, beakers, pipettes, burettes, flasks, or bottles, you are being asked to
measure volume. Volume measurements in the International System of Units use the metric system rather
than the imperial system in order to standardize these measurements around the globe. Thus, for volume
measurements, we use liters (L) for large volumes and milliliters (mL) for smaller volumes measured in the
lab. Look at the figure below and determine what volumes are present in each piece of equipment.

Volume equipment pieces. (Source: Richard Parsons. CC-BY-SA)

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Graduated Cylinder Graduated Pipet

The contrary is also true. What if you were to measure out 5 g of a solid, or 3 cm of wire, or the temperature
of a solution; would any of the objects in Figure 5 be helpful? Why not? These objects are not helpful
because these units of measurement are not volumes and all of these pieces of equipment measure volume.
For the measurements you need to take, you would need different pieces of equipment. Look at the figure
below and match the three required measurements with the pieces of equipment shown.

a) 5 g of a solid

b) 3 cm of wire

c) temperature of a solution

Examples of measuring devices. (Source: http : //en . wikipedia . org/wiki/ Image : Steel_ruler_closeup .
jpg http : //en . wikipedia . org/wiki/File : Clinical_thermometer_38 . 7 . JPG http : //en . wikipedia .
org/wiki /Image: Digi-keukenweegschaall284.jpg. GNU-FDL, GNU-FDL, GNU-FDL)

Equipment Determines the Significant Figures

In the previous section, we looked at a lot of equipment that is used for measuring specific units. The
graduated cylinder that measures volume, the balance that measures mass, and the thermometer that
measures temperature are a few that we looked at before. We also saw that of two types of balances,
one type of balance can more precisely measure mass than the other. The difference between these two
balances has to do with the number of significant digits that the balances are able to measure. Remember
the pan balance could measure to +/ - 0.1 g and the digital balance can measure to +/ - 0.01 g.

Before going any further, what do we recall about significant digits? A measurement can only be as
accurate as the instrument that produced it. A scientist must be able to express the accuracy of a number,
not just its numerical value.

The instruments that we choose for the laboratory experiments depend on the required amount of accuracy.

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For example, if you were to make a cup of hot chocolate at home using powdered cocoa, you would probably
use a measuring spoon or a teaspoon. Compare this to the requirement of massing out 4.025 g of sodium
bicarbonate for a reaction sequence you are doing in the lab. Would the teaspoon do? Probably not! You
would need to have what is known as an analytical balance that measures to +/ - 0.001 g.

Accuracy and Precision

Accuracy and precision are two words that we hear a lot in science, in math, and in other everyday events.
They are also, surprisingly, two words that are often misused. How often have you heard these terms?
For example, you often hear car advertisements that talk about their precision driving ability. But what
do these two words mean. Accuracy is how close a number is to the actual or predicted value. If the
weatherperson predicts that the temperature on July 1 st will be 30°C and it is actually 29°C, she is likely
to be considered pretty accurate for that day.

Once you have gone into the lab and made measurements, whether they are mass, volume, or length, how
do you know if they are correct? Accuracy is the difference between a measured value and the accepted
- or what we call the correct - value for that quantity. To improve accuracy, scientists will repeat the
measurement as many times as is possible. Precision is a measure of how close all of these measurements
are to each other. Therefore, measurements can have precision but not very close accuracy. An example of
accuracy of measurements is having the following data: 26 mL, 26.1 mL, and 25.9 mL when the accepted
value is 26.0 mL. This data also shows precision. However, if the data had been 25.2 mL, 25.0 mL, and
25.2 mL, they would show precision without accuracy.

Sample question: Jack collected the following volumes when doing a titration experiment: 34.25 mL,
34.30 mL, 34.60 mL, 34.00 mL, and 34.50 mL. The actual volume for the titration required to neutralize
the acid was 34.50 mL. Would you say that Jack's data was accurate? Precise? Both accurate and precise?
Neither accurate nor precise? Explain.

Solution: All of Jack's data would be accurate because they are close to the true value of 34.50 mL. The
data would also be precise having only 2% variance between the highest number and the lowest number.

The distinction between accuracy and precision and its importance in science is demonstrated in an An-
nenberg video at Video on Demand - The World of Chemistry - Measurement (http : //www. learner .
org/resources/series61 .html?pop=yes&pid=793#)

Lesson Summary

• The task in the experiment determines the unit of measurement; this then determines the piece of
equipment. Example: If mass is to be measured, a balance will be chosen as the piece of equipment.

• Conversely, the piece of equipment chosen will determine the unit of measurement. Example: If a
graduated cylinder is chosen, the unit of measurement will be volume (mL or L).

• Each piece of equipment has a specified number of significant digits to which it is able to measure.
Example: A household thermometer may measure to ± VC or ± 1°F, where as an ordinary high
school alcohol thermometer measures to ± 0.1°C.

• Significant digits are used in all parts of quantitative measurements in science. Five main rules are
provided to read the significant digits of numbers and two main rules for solving algebraic equations
maintaining proper significant digits.

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• Accuracy is how close the value is to the actual value (remember A and a).

• Precision is how close values are in an experiment to each other. Precision is dependent on the
significant digits of the instrument or measurement.

Review Questions

1. Suppose you want to hit the center of this circle with a paint ball gun. Which of the following are
considered accurate? Precise? Both? Neither?

2. Four students take measurements to determine the volume of a cube. Their results are 15.32 cm 3 ,
15.33 cm 3 , 15.33 cm 3 , and 15.31 cm 3 . The actual volume of the cube is 16.12 cm 3 . What statement(s)
can you make about the accuracy and precision in their measurements?

3. Find the value of each of the following to the correct number of significant digits.

(a)

1.25 + 11

(b)

2.308

-1.9

(c)

498-

97.6

(d)

101.3

-=-12

(e)

25.69

xO.51

4. Why is the metric system used in chemistry?

5. Distinguish between accuracy and precision.

6. How many significant digits are present in each of the following numbers:

(a) 0.002340

(b) 2.0X10- 2

(d) 3.00 x10 s

7. Nisi was asked the following question on her lab exam. When doing an experiment, what term best
describes the reproducibility in your results? What should she answer?

(a) accuracy

(b) care

(d) significance

(e) uncertainty

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8. Karen was working in the lab doing reactions involving mass. She needed to weigh our 1.50 g of each
reactant and put them together in her flask. She recorded her data in her data table and began to
look at it (Table 3.4). What can you conclude by looking at Karen's data?

(a) The data is accurate but not precise.

(b) The data is precise but not accurate.

(d) The data is precise and accurate.

(e) You really need to see the balance Karen used.

Table 3.4:

Mass of Reactant 1 Mass of Reactant 2

Trial 1 1.45 + 0.02 g 1.46 + 0.02 g

Trial 2 1.43 ± 0.02 g 1.46 + 0.02 g

Trial 3 1.46 + 0.02 g 1.50 ± 0.02 g

9. Find the value of each of the following to the correct number of significant digits.

(a) 3.567 + 3.45

(b) 298.968 + 101.03

(d) 27^5.67

(e) 423x0.1

Further Reading / Supplemental Links

The website listed below offers lessons, worksheets, and quizzes on many topics in high school chemistry.

• Lesson 2-2 is on Accuracy and Precision, http : //www . f ordhamprep . org/gcurran/sho/sho/lessons/
lessonl2.htm

• http : //learner . org/resources/series61 . html

The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos.
You are required to register before you can watch the videos but there is no charge. The website has two
videos that apply to this lesson. One is a video called Measurement, The Foundation of Chemistry
that details the value of accuracy and precision. Another video called Modeling the Unseen relates to
the scientific method and the use of models.

Vocabulary

significant digits A way to describe the accuracy or precision of an instrument or measurement.

accuracy How close a number is to the actual or predicted value.

precision How close values are in an experiment to each other.
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Labs and Demonstrations for Making Measurements
Teacher's Pages for Thermometer Calibration

Investigation and Experimentation Objectives

In this activity, the student will determine one case of experimental error and recognize the limitations of
measuring instruments.

Notes:

Thermometers should be stored vertically when not in use. You can stand them upright in a large beaker
or in tall test tube racks. When thermometers are stored horizontally, they sometimes suffer a separation
of the liquid near the top. If you have thermometers with separated liquid, you can sometimes shake them
down or bounce the bulb gently on a folded towel to rejoin the liquid.

Answers to Pre-Lab Questions

1. Why is a mixture of ice and water, rather than ice alone, used in calibrating a thermometer?

You cannot be sure of the temperature of solid ice. It might be — 10°C When ice and water are both
present and in equilibrium, you can be sure the temperature of the mixture is 0°C.

2. Why does the boiling point of a liquid vary with the barometric pressure?

Water boils when its vapor pressure is equal to the surrounding pressure. If the surrounding pressure is
above or below normal atmospheric pressure, then the boiling point of a liquid will be above or below its
normal boiling point.

3. What is the approximate boiling point of pure water at 380 Torr?
Between 80° C and 82° C.

4. What is the approximate boiling point of pure water at 800 Torr?
Near 101°C

5. Food products such as cake mixes often list special directions for cooking the products in high altitude
areas. Why are special directions needed? Would a food product needing such directions require a longer
or shorter time period to cook under such conditions?

At high altitudes, the atmospheric pressure is less than normal atmospheric pressure and therefore, the
boiling point of water is below 100°C Since boiling water is less than 100°C, cooking in boiling water will
take longer.

Table 3.5: Vapor Pressure of Water at Various Temperatures

Temperature in °C Vapor Pressure in mm Temperature in °C Vapor Pressure in mm

of Hg of Hg

-10 2.1 52 102.1

-5 3.2 54 112.5

4.6 56 126.8

2 5.3 58 136.1

4 6.1 60 149.4

6 7.0 62 163.8

8 8.0 64 179.3

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Table 3.5: (continued)

Temperature in °C

Vapor Pressure in mm
of Hg

Temperature in °C

Vapor Pressure in mm
of Hg

12
14
16

18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50

9.2

10.5

12.0

13.6

15.5

17.5

19.8

22.4

25.2

28.3

31.8

35.7

39.9

44.6

49.7

55.3

61.5

68.3

75.7

83.7

92.5

100

102

104

106

196.1
214.2
233.7
254.6
277.2
301.4
327.3
355.1
384.9
416.8
450.9
487.1
525.8
567.0
610.9
657.6
707.3
760.0
815.9
875.1
937.9

Thermometer Calibration

Background Information

The most common device for measuring temperature is the thermometer. The typical thermometer used in
general chemistry labs has a range from -20°C to 120°C Most laboratory thermometers are constructed of
glass and therefore are very fragile. Older thermometers contain mercury as the temperature sensing liquid
while newer thermometers contain a red colored fluid. The mercury thermometers are hazardous if they
break because mercury vapors are poisonous over long periods of inhalation and the mercury vaporizes
slowly and so when it is spilled, the lab is toxic for several months unless every drop of mercury is picked
up. The red colored liquid thermometers are also hazardous if they break because the liquid is flammable
and may be toxic. Great care should be exercised when handling thermometers of either kind.

The typical laboratory thermometer contains a bulb (reservoir) of temperature sensing fluid at the bottom;
it is this portion of the thermometer which actually senses the temperature. The glass barrel of the
thermometer above the liquid bulb contains a fine capillary opening in its center, into which the liquid
rises as it expands in volume when heated. The capillary tube in the barrel is very uniform in its cross-
section all along the length of the thermometer. This insures that the fluid will rise and fall uniformly
when heated or cooled.

(NOTE: laboratory thermometers look like clinical thermometers for taking people's temperatures but they
are not the same. The clinical thermometer has a constriction in the tube so that after the temperature goes
up and the thermometer is removed from the heat source, the liquid will not go back down. Such clinical
thermometers must be shaken to lower the temperature reading before each use. Lab thermometers have

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no such constriction and hence the temperature reading immediately starts down when the heat source
is removed. For that reason, lab thermometers must be read while the bulb is still in contact with the
material whose temperature is being taken.)

Because thermometers are so fragile, it is a good idea to check them, now and then, to make sure they
are still working properly. To check a thermometer, a process of calibration is used. To do this, you
will determine the reading given by your thermometer in two systems whose temperature is known with
certainty. If the readings of your thermometer differ by more than one degree from the true temperatures,
it should be removed from use.

A mixture of ice and water which has reached equilibrium has a temperature of exactly 0°C and will be used
as the first calibration point. The second calibration point will be boiling water whose exact temperature
must be determined using the barometric pressure in the lab.

Pre-Lab Questions

1. Why is a mixture of ice and water, rather than ice alone, used in calibrating a thermometer?

2. Why does the boiling point of a liquid vary with the barometric pressure?

3. What is the boiling point of pure water at 380 Torr?

4. What is the boiling point of pure water at 800 Torr?

5. Food products such as cake mixes often list special directions for cooking the products in high altitude
areas. Why are special directions needed? Would a food product needing such directions require a
longer or shorter time period to cook under such conditions?

Purpose

In this experiment, you will check a thermometer for errors by determining the temperature of two stable
equilibrium systems.

Apparatus and Materials

• Thermometer

• 400 mL beaker

• 250 mL beaker

• distilled water

• ice

• hot plate

• stirring rod

• boiling chips.

Safety Issues

Mercury thermometers are hazardous if they break because mercury vapors are poisonous over long periods
of inhalation and the mercury vaporizes slowly and so when it is spilled, the lab is toxic for several months
unless every drop of mercury is picked up. The red colored liquid thermometers are also hazardous if they
break because the liquid is flammable and may be toxic. Great care should be exercised when handling
thermometers of either kind.

Procedure

Fill a 400 mL beaker with ice and add tap water until the ice is covered with water. Stir the mixture is a
stirring rod for one minute. Dip the thermometer into the ice water mixture so that the thermometer bulb
is approximately centered in the mixture (not near the bottom or sides). Leave the thermometer in the
mixture for two minutes and then read the thermometer to the nearest 0.2 degree while the thermometer
is still in the ice water bath. Record the temperature.

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Allow the thermometer to return to room temperature by resting it is a safe place on the laboratory table.

Half fill a 250 mL beaker with distilled water and place it on a hot plate. Add 2 or 3 boiling chips
to the water. Heat the water to boiling. Dip the thermometer into the boiling water making sure the
thermometer does not get near the bottom, sides, or top of the water. Hold it there for 2 minutes and
record the temperature reading to the nearest 0.2 degree.

Ask your instructor for the current barometric pressure reading in the laboratory room, look up the actual
boiling point of water at this pressure and record.

Data

Actual freezing point of water =

Freezing point determined by your thermometer =

Difference between correct and trial values =

Barometric pressure in the room =

Actual boiling point of water at this pressure =

Boiling point determined by your thermometer =

Difference between correct and trial values =

Post-Lab Questions

1. Calculate the percent error of your measurement of the freezing point of water.

actual value — trial value

b error

actual value

xlOO

2. Calculate the percent error of your measurement of the boiling point of water.

_. actual value - trial value

% error = X 100 =

actual value

Table 3.6: Actual Boiling Point of Water versus Various Room Pressures

Room Pressure (mm of Hg)

Boiling Point of Water (°C)

750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766

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99.6

99.7

99.8

99.9

100.0

100.1

100.2

Table 3.6: (continued)

Room Pressure (mm of Hg) Boiling Point of Water (°C)

767 100.2

768 100.3

769 100.3

770 100.3

3.3 Using Data

Lesson Objectives

• Recognize patterns in data from a table of values, pictures, charts and graphs.

• Make calculations using formulae for slope and other formulae from prior knowledge.

• Construct graphs for straight lines.

• Construct graphs for curves.

• Read graphs using the slope of the line or the tangent of the line.

Introduction

Earlier, we learned about qualitative and quantitative observations, and that with quantitative observa-
tions, we need measurements. In science, measurements mean data. In this upcoming section, we will
delve deeper with data to look at patterns and to graph data. Sometimes we can graph the data, make
calculations, sketch the line, and calculate the slope. All of these quantitative observations help us to
formulate a conclusion that will be based on evidence.

Recognizing Patterns in the Data

As stated earlier, data can provide enormous information to scientists for making interpretations and
drawing conclusions. In order for scientists to do this, they have to be able to look at a set of data
and recognize patterns. Data can be in the form of pictures, charts, or graphs. Take for example, the
picture found in (Figure 3.1). This, although not particularly chemistry related, has much to do with the
concept of pattern recognition and the data gathered from these patterns. What do you believe scientists
determined when first viewing this image prior to August 23, 2005?

Now, let us look at a common chemistry example. We all know that metals are supposed to be chemically
reactive. How do we know that? Well, that is a property of metals. But how reactive are they? How
can you tell? What do you know about the periodic table that would help you determine this right now?
Look at Table 3.7 and see if you can gather a little evidence that can solidify your conclusions about the
chemical reactivity of metals and the periodic table.

Table 3.7: Information about Metals and Reactivity

Metal Chemical Reactivity in Atmosphere

Sodium Stored in toluene, extremely reactive

Potassium Burns in 02 in seconds

Calcium Slower to react with Oi than its neighbors to the

right

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Table 3.7: (continued)

Metal

Chemical Reactivity in Atmosphere

Titanium

Aluminum

Gold

Platinum

Copper

Iron

Resists corrosion after forming an oxide barrier

Does not react with oxygen

Does not react with H2O, will react slowly with O2

Rusts in 02

What kinds of conclusions can you make from reading this table? Can you determine that the reactions
between the metals and the oxygen in the air decrease going across a row on the periodic table? Did you
notice that the alkali (group one) metals (sodium and potassium) are the most reactive of all the metals
with oxygen and the alkaline earth (group two) metals (calcium) are the second most reactive? Yes, they
are. Aluminum as well as titanium will actually become coated with an oxide of their metals, which acts
like a protective shield against further reaction. Look at the figure below. The copper has turned green as
a result of this protective coating that formed on the copper rooftop. Have you ever seen this before?

The current Hotel Vancouver took over a decade to build during the 1930s as the Great Depression
put a temporary halt to construction. It was Vancouver's tallest building from 1939 to 1972. (Source:
http : //commons . wikimedia . org/wiki/File : HotelVancouver . jpg . CC-BY-SA)

What kind of observations are these? Are they quantitative? No, of course not because they have no
measurements attached; they are qualitative observations.

What about other types of data that involve quantitative observations? Can we look at measurements and
determine patterns? Yes we can. Statisticians, weather persons, stock market workers, sports analysts,
and chemists (to name a few occupations) do this on a daily basis for their regular jobs. What if you were
trying to identify an unknown substance based on a volume displacement experiment. You were given a
series of known substances of known masses, and you then determined how much volume they displaced
in a cylinder of water. Through your experiment, the following data was recorded (Table 3.8).

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Figure 3.1: Hurricane Katrina taken on Aug. 28, 2005, at 11:45 a.m. EDT by NOAA when the storm was
a Category Five hurricane.

Table 3.8: Volume Displacement Data

Substance

Mass

Volume ol Water Displaced

Aluminum
Iron
Copper
Silver
Zinc
Lead

2.7 g

7.86 g
8.92 g
10.5 g
7.14 g
11.34 g

1.0 mL

1.0 mL
1.0 mL
1.0 mL
1.0 mL
1.0 mL

First, which metal do you think is the densest, just looking at the data table? Do you think it would be
lead or aluminum? How can you tell? Lead is the right answer because there is a heavier mass of lead
displacing the same volume of water. Did you notice that copper is heavier than iron, and iron is heavier
than zinc? Now what kind of data do you think this is? Is it qualitative or quantitative? Quantitative is
the right answer because we are dealing with measurements.

An interesting note is that 19.3 g of gold would have displaced the same volume of water. Would it be the
densest? Yes it would. It also means that this same amount of gold converts to approximately 212.5 lb/gal.
Remember seeing in television or movies where the villain is running off with a bag full of gold bars? This
would be rather difficult knowing the density of the metal from this table. Being able to read tables and
data gives us the power to understand the world around us.

Another way to generalize trends in data and to make interpretations based on these trends is to plot a
graph. Graphs are like numerical pictures that provide an image of the data collected in an experiment.
Suppose you were asked to record the temperature of a mixture as it was slowly heated in a hot water
bath. You record the following data as you watch your experiment (Table 3.9). You then plot the data
on a graph to see what the numbers tell you.

Table 3.9: Time vs. Temperature

Time (s)

Temperature (°C)

1
2
3

5
6

7
8

23.5
24
25
26
27
9128
29
30
31

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Temperature

-I — h

Time (s)

1 1

Looking at the graph, what do you notice about the temperature of the mixture as it is heated in the hot
water bath? It is constantly increasing. What does the initial point, or the y— intercept represent? It is the
value of the initial temperature, most likely room temperature. Using a graph, it is clear to see that the
y—axis, in this case temperature, is dependent on the x-axis. For our example, the independent variable
is time. What does this mean? It means that for a change to be made in temperature, time must pass.

Being able to plot tables of values and read the corresponding graphs is an important skill, not only for
mathematics but also for science. Interpretations can be made by using either of these representations; one
may be more visual and thus sometimes easier to interpret than the other. Now try one example where
you have to plot a graph and make some interpretations.

Sample Question: The following data represents the marks of 12 students in a Math Test and in a Chemistry
Test. All Marks are out of 50.

Math Marks vs. Chemistry Marks

Math

Chemistry

a. Plot this data on an x - y axis, with Math marks on the x-axis.

b. Draw the line of best fit. A line of best fit is drawn on a scatter plot so that it joins as many points
as possible and shows the general direction of the data. When constructing the line of best fit, it is also
important to keep, approximately, an equal number of points above and below the line.

c. Estimate the chemistry mark of a student who scored 32 on a math test.

d. Estimate the math mark of a student who scored 45 on a chemistry test.

e. Based on the trend found in the data, what can you say about the relationship between math and
chemistry marks?

Solution: a) and b)

Graph 2: Math Marks vs. Chemistry Marks

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c) The chemistry mark when a student makes 32 in math is approximately 25. We can see this by the
blue line. By interpolating the data, we can draw a line up from the 32 mark on the x-axis to where the
line of best fit runs through the data points. Here, we cut across to the y-axis to find the corresponding
chemistry mark.

d) The math mark when a student makes 45 in chemistry is approximately 47. We can see this by the
green line. By interpolating the data, we can draw a line across from the 45 mark on the y-axis to where
the line of best fit runs through the data points. Here, we draw a line down to the x-axis to find the
corresponding math mark.

e) The trend shows that as math marks increase, so do chemistry marks. The dependent variable in this
graph is the chemistry mark.

Making Calculations With Data

Being able to recognize patterns from tables, charts, pictures, and graphs is a worthy skill for any scientist
and science student. By having charts, pictures, tables, and graphs, you can also perform a large variety
of calculations depending on the independent and dependent variables. From here we can accomplish such
things as making further predictions, drawing more conclusions, or identifying unknowns. Let's say, for
example, in the density experiment from earlier, we were given an unknown for our experiment. Now, take
the same chart but add a fourth column representing the density of the metal (Table 3.10). Recall the
formula for density:

Density

mass
volume

Now, filling in the fourth column, using the density formula for our experimental data, we see the following
information. Remember density = mass/volume or, in the table below, column 4 = column 2/ column 3.

Table 3.10:

Substance

Mass

Volume of Water Dis- Density
placed

Aluminum

2.7 g

1.0 mL

2.7 g/mL

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Table 3.10: (continued)

Substance Mass Volume of Water Dis- Density

placed

Iron 7.86 g 1.0 mL 7.86 g/mL

Copper 8.92 g 1.0 mL 8.92 g/mL

Silver 10.5 g 1.0 mL 10.5 g/mL

Zinc 7.14 g 1.0 mL 7.14 g/mL

Lead 11.34 g 1.0 mL 11.34 g/mL

Unknown 1.78 g 0.2 mL 8.9 g/mL

Just by doing this calculation, we can identify our unknown in the experiment? Sure, it is copper. Calcula-
tions are frequent in chemistry, as you will learn. Most times you will see we will use a variety of formulae
to solve problems similar to those you would solve in any course. The relationship from comparing sets of
answers give us the interesting parts of these types of calculations.

Sample Question: How long does it take you to run 3.5 miles at 7 mph?

Solution: We know that: speed = di t s ^ ce

Therefore:

distance
time

time

speed
3.5 mi

7 mi/h
time = 0.5 h

Another important skill with calculations is converting from one unit to another. Frequently in chemistry,
data will appear in problems that require us to use conversion factors before completing the problem. Some
conversion factors include 100 cm = 1 m, 1000 mL = 1 L, and 1 km = 1000 m.

Sample Question: The speed of light is 3.00 X 10 8 m/s. The speed of sound is 1230 km/h. How much
faster is the speed of light than the speed of sound.

Solution: Speed of Sound = 1230 km/h x ±f2Lm x _L^_ x |>5i

Speed of Sound = 342 m/s

speed of light _ 3.00 x 10 8 m/s

speed of sound 342m/s

speed of light _ ^

B - 8.77 x 10°

speed of sound

In other words, the speed of light is almost 900, 000 times faster than the speed of sound. Amazing isn't
it!

Preparing Graphs From Data

Most times in the laboratory, we collect data of some sort and then carry it back to our desk to analyze. We
want to determine the melting point of an unknown solid, so we take the melting points of various knowns
and then that of our unknown; following this, we make a table, finally writing the data gathered from our

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experiment into the table. Some laboratory experiments that we, as scientists, do require us to draw graphs
in order to interpret the results and make any conclusions. Drawing a graph that anyone can understand
is a useful skill to any scientist. Graphs have to be properly labeled on the x-axis (the horizontal) and the
y-axis (the vertical). The graphs should indicate a straight line or smooth curve indicating that the data
is continuous. A straight line represents a linear relationship; a curved line does not.

Sample Question: Medical practitioners have been studying the heart for a long time. As a result, we can
now calculate your heart rate based on a formula derived from your age. Plot the table of values given
below. Properly label the graph include the independent and dependent variable. Draw the line all the
way to the y-axis so that you can find the y- intercept. Finally, find your age on the x-axis and then find
your maximum heart rate by drawing a vertical line up to the graph.

Age in Years (x) 30 40 50 60

Beats Per Minute (y) (maximum heart rate) 90 80 70 60

Solution: At the age of 17, the maximum heart rate (beats per minute) read from the graph is 103.
Heart Rate vs Age

Maximum Heart Rate
(beats per minute)
Dependent

Age (Years)
Independent

Reading Results From the Graph

Many properties in chemistry lead to linear relationships when plotted. We saw this with the temper-
ature/time relationship. Other properties in chemistry do not form this linear relationship. Take for
example the relationship between concentration and temperature. Remember the last time you made a
cup of instant coffee or hot chocolate? Why did you boil the water? What would have happened if you used
warm tap water or even cold water from the refrigerator? Putting aside the anticipated taste difference,
what would have happened to the solid you were trying to dissolve? The amount of instant coffee or hot
chocolate powder in your cup that actually dissolves in warm or cold water would be small compared to
when you use boiling water. This property is known as solubility. The solubility of a substance is the
amount that can dissolve in a given amount of solution. Solubility is affected by the temperature but rarely
linearly. Look at the data table below (Table 3.11). This data is for the solubility of KCIO3 (potassium
chlorate) in water.

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Table 3.11:

Temperature (°C)

Solubility ( g/100 mL H 2 0)

100

3.3

7.3

13.9

23.8

37.5

56.3

Now graph the data and see what kind of curve we get.
Solubility of Potassium Chlorate

Notice how when the line is drawn, the relationship between grams of potassium chlorate that dissolve
in 100 mL of water and temperature is not linear but curved. We can still interpret the data as we did
earlier. For example, what is the solubility of KCIO3 (how much KCIO3) when the temperature is 75°C or
at room temperature.

Solubility of Potassium Chlorate

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Using the same procedure, we draw a line up from 75°C and then over to the v-axis. It reads 34.5 g/100 mL.
Therefore, 34.5 g of KCIO3 can dissolve in 100 mL of H2O at 75° C.

Now you try one:

Sample Question:

Ammonia, NH3, and sodium chloride, NaCl, are known to have the following solubility data (Table 3.12).

Table 3.12:

Temperature (°C)

NH 3 Solubility ( g/100 mL H 2 0) NaCl Solubility ( g/100 mL H 2 0)

100

88.5
56.0
34.0
20.0
11.0
7.0

35.7
35.9
36.4
37.1
38.0
39.2

a) Properly graph the data for each substance.

b) Are either of these linear? Explain.

c) What would be the solubility of each of the substances at 50°C?
Solution:

Solubility of NH 3 and NaCl

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70
60

Solubility
in water 50
(g/mL)

N'S-

S 10 15 20 25 30 35 40 45 50 55 60 65 70 75 BO 65 90 95 100

Temperature ( C)

b) The Solubility data for NaCl represents a linear relationship when graphed. You can see this with the
blue line in the graph above. The NH% line (in red) is curved; therefore it is non-linear.

c) The solubility of NH 3 at 50°C is 27 g/100 mL H 2 0. The solubility for NaCl at 50°C is 37.5 g/100 niL H 2 0.
The graph below is marked to show the line travelling up from 50°C and over to the y-axis (solubility) to
find the answer for both of these parts.

Solubility of NH 3 and NaCl

70
60

Solubility
in water 50
(g/mL)

37.5
30

"--■-,

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75

Temperature { C)

30 B5 90 95 100

We can do more than just graph and read graphs of linear and non-linear data in order to make conclusions.
For this, we use formulas like the slope of a line. Remember slope from math class? It is a formula used
to find the rate at which one factor is affecting the other, either positively or negatively. Remember the
formula for slope from math class?

rise y 2 - yi

slope = or m =

run X2 - x\

Let's look at how the slope formula can be used on a graph to see how one factor is affecting another in
an experiment.

A modern hot air balloon. (Source:http : //en . wikipedia . org/wiki/ Image : 2006_0 j iya_balloon_f est ival_
011.jpg- CC-BY-SA)

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In the 18005, the use of the hot air balloons was extremely popular as a sport as well as an extracurricular
activity for those who could afford the luxury. Up to this point, the study of the relationship of gases and
the factors of temperature, pressure, and volume was limited to Robert Boyle's experiments with pressure
and volume. Jacques Charles came along with his experiments on the relationship between volume and
temperature. Here is some typical data from a volume/temperature experiment with gases (Table 3.13).

Table 3.13:

Temperature (C°0)

Volume of Gas (cm 3

100

120

60
65

75
80
85

Volume of a Gas vs. Temperature

I I I I I I II

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Now, find out what affect the temperature has on the volume of the gas. In other words, find the slope.
Pick two points that are on the line and use the equation above to find the value of m.

y2-

X2 -

80-

-65

100

-40

" 60

= 0.25

cm 3 /'

What does this mean? It means that for each increase in temperature of 1°C, the volume increased by
0.25 cm 3 (or 0.25 mL). This translates to approximately 1 mL increase every 4°C. This is a positive
increase (notice the slope is increasing or going up). Now let's try another one.

Look at Table 3.14 for a set of data from an experiment performed between bromine and formic acid in
a laboratory setting. The reaction was performed to see if the decrease in bromine concentration could
cause the reaction to subsequently slow down. In other words, if they took some of the bromine out of the
reaction, would the reaction start to slow down? Look at the data and see what happened.

Table 3.14:

Reading Concentration of Bromine Time

(mol/L)

1 0.1

2 0.07 0.75

3 0.05 1.75

4 0.035 2.49

5 0.02 3.48

6 0.01 5

7 0.005 6.2

8 0.001 7.5

9 0.0 8.8

10 0.0 9

Look at the data table, can you tell if an increase in bromine concentration had an effect on the rate of a
reaction? Did it make it go faster or slower. Let's take a look. From 7\ to T-j the bromine concentration
decreased from 0.1 mol/L to 0.005 mol/L, a decrease of 0.095 mol/L. The time it took for this decrease
was 6.2 seconds. What does this tell us about the rate? A preliminary conclusion would be that a decrease
in bromine concentration causes the rate of the reaction to also decrease. A graph might make it a little
easier to make conclusions based on the data.

Concentration of Bromine vs. Time
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Using slope to calculate the effect here is not as easy because we have a curve. We can easily see that as
the concentration increases, so does the rate, but by how much? The slope would actually tell us that. In
order to determine the slope of a curve, you need to draw in a tangent to the curve. A tangent is just a
straight line drawn to the curve; from this you would calculate the slope. Look at the graph below, the
tangents are drawn in using a red pen.

Concentration of Bromine vs. Time

If we draw a tangent, look, we have a straight line. We can now, find two points.

y2-yi

x 2 -xi
0.06 - 0.07

1.1-0.75
m = -0.028 mol/L • s

This means that as the concentration of bromine decreases, so does the rate of the reaction. Look at the
units for the slope. The units are mol/L s. These are the units for rate. This means, interestingly enough,
that as the concentration goes down, the reaction slows down.

101

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Lesson Summary

• Patterns can be found in data sets, pictures, charts, and graphs. From here scientists can make
interpretations of the data and draw conclusions.

• When drawing graphs of tables of values, a straight line or a smooth curves can be drawn. Some
data sets do require a line of best fit.

• A line of best fit is drawn on a scatter plot so that it joins as many points as possible and shows the
general direction of the data. When constructing the line of best fit, it is also important to keep,
approximately, an equal number of points above and below the line.

• Conversion factors are necessary for calculations where the units do not match. For example, km
and m.

• Recall the slope formula: m = } ' 2 yi

• For curved lines, remember to draw the tangent first and then find the slope of the tangent line.

Review Questions

1. Why is the slope of a graph so important to chemistry?

2. What would you do to find the slope of a curved line?

3. What is a conversion factor used for?

4. Of the following professions, choose the one that uses data to find the identity of unknown finger-
prints?

(a) analytical chemist

(b) archaeological chemist

(d) forensic scientist

(e) quality control chemist

5. Which speed is the slowest?

(a) 200 m/min (200 m/min X 1 km/1000 m = 0.2 km/min)

(b) 0.2 km/min

(d) 1.0 X 10 5 mm/min (1.0 X 10 5 mm/min X 1 m/1000 mm X 1 km/1000 m = 0.1 km/min)

(e) 10 mi/h (10 mi/h X 1.603 km/mi X lh/60 min = 0.267 km/min or 0.27 km/min)

6. Andrew was completing his density lab for his chemistry lab exam. He collected the following data
in his data table (Table 3.15).

(a) Draw a graph to represent the data.

(b) Calculate the slope.

(d) Can you help Andrew determine what his unknown is by looking in a standards table?

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Table 3.15:

Mass of Solid (g) Volume of Solution (niL)

3.4 0.3

6.8 0.6

10.2 0.9

21.55 1.9

32.89 2.9

44.23 3.9

55.57 4.9

7. Donna is completing the last step in her experiment to find the effect of the concentration of ammonia
on the reaction. She has collected the following data from her time trials and is ready for the analysis
(Table 3.16). Donna is now required to graph the data, describe the relationship, find the slope and
then discuss the meaning of the slope. Help Donna with the interpretation of her data.

Table 3.16:

Time (s) Concentration (mol/L)

0.20 49.92

0.40 39.80

0.60 29.67

0.81 20.43

1.08 14.39

1.30 10.84

1.53 5.86

2.00 1.95

2.21 1.07

2.40 0.71

2.60 0.71

Further Reading / Supplemental Links

• http : //en . wikipedia . org/wiki/Metal

• http : //en . wikipedia . org/wiki/Hurricane_Katrina

• http : //en . wikipedia . org/wiki/Solubility_table

Vocabulary

chemical reactivity An observation of the behavior of the element of compound based on its position
in a reactivity (or activity) series.

periodic table An arrangement of elements in order of increasing atomic number.

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alkali metals Group 1 metals of the periodic table (H,Li,Na,K,Rb,Cs,Fr).

alkaline earth metals Group 2 metals of the periodic table (Be,Mg,Ca,Sr,Ba,Ra).

density Measurement of a mass per unit volume. Density = rpr^- .

graphs Pictorial representation of patterns using a coordinate system (x — y axis).

dependent variable The variable that changes depending on another variable (y-axis variable).

independent variable The variable that changes to cause another variable to change (x-axis variable).

y-intercept Where the line crosses the y-axis.

conversion factor A ratio used to convert one unit to another.

linear relationship A relationship where the x-values change proportionally with the y-values leading
to a straight line.

non-Linear relationship A relationship where the x-values do not change proportionally with the y-
values leading to a curved line.

a line of best fit A line drawn on a scatter plot so that it joins as many points as possible and shows
the general direction of the data. When constructing the line of best fit, it is also important to keep,
approximately, an equal number of points above and below the line.

slope A formula to find the rate at which one factor is affecting the other.

y2-vi
m =

X2 - Xl

tangent A straight line drawn to the curve.

solubility The amount of a substance that can dissolve in a given amount of solution.

Labs and Demonstrations for Using Data
Density Determination

Investigation and Experimentation Objectives

In this activity, the student will make and record observations using rulers, balances, and graduated
cylinders. The student will use his measurements to mathematically calculate the density of both regular
and irregular objects.

Pre-Lab Discussion

Density is defined as the mass per unit volume of a substance. The table below lists the densities of some
well known substances.

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Table 3.17: Density of Some Common Substances

Substance

Density

Substance

Density

Water

1.0 g/cm 3

Aluminum

Oxygen gas

0.0013 g/cm 3

Iron

Sugar

1.6 g/cm 3

Lead

Table salt

2.2 g/cm 3

Gold

2.7 g/cm 3
8.9 g/cm 3
11.3 g/cm 3
19.3 g/cm 3

Density

mass in grams
volume in mL

Example: Calculate the density of a piece of lead whose mass is 226 grams and whose volume is 20.0 mL.
Also calculate the density of a sample of gold whose mass is 57.9 grams and whose volume is 3.00 mL.

Solution:

Density of Lead
Density of Gold

mass 226 g

volume 20.0 mL

mass 57.9 g

volume 3.00 mL

11.3 g/mL
19.3 g/mL

Methods of Measuring Mass and Volume

CUBE

SPHERE

CYLINDER

V-OOWW

CZZ2

y= 7ir 2 h

105

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Procedure:

1. Obtain a regularly shaped object from your teacher. Measure and record its mass in grams and its
dimensions in centimeters.

3. Measure and record the mass of a clean, dry, 10 - mL graduated cylinder.

4. Add exactly 10.0 mL of distilled water to the cylinder. Measure and record the combined mass of the
cylinder and water.

5. Repeat steps 3 and 4 with glycerol instead of water.

Data Table
Object

code name or letters

width

height

length

volume of water before block

volume of water after block

Distilled Water

volume

mass of empty graduate

combined mass

Glycerol
volume

mass of empty graduate

combined mass

Calculations

1. Find the volume of the solid object using the dimensions and appropriate formula.

2. Find the volume of the solid object using water displacement.

3. Find the density of the block using the volume from calculation 1.

4. Find the density of the block using the volume from calculation 2.

5. Find the density of the distilled water.

6. Find the density of the glycerol.

7. If your teacher gives you the accepted values for the densities in this lab, calculate the percent error
for your values.

(experimental value) — (actual value)

% error = — X 100

(actual value)

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Demonstration of Density of Diet Soda vs. Regular Soda

Investigation and Experimentation Objectives

In this activity, the student will make observations and use critical thinking to suggest a possible explanation
for the observations.

Brief description of demonstration

A can of diet soda and regular soda are place into a clear container of water. The diet soda floats while
the regular soda sinks.

Materials

• 12 oz. can of diet soda

• 12 oz. can of regular soda, preferably the same brand

• Clear container with enough volume so the can has room to sink totally

Procedure

Fill the clear container to within 5 cm of the top with water. Place the diet soda into the container. It will
float. Place the regular soda into the container. It will sink.

Hazards

None.

Disposal

Pour the water down the sink.

Discussion

This is a good demonstration of density and for the discussion of dependent and independent variables.
The only significant difference between the cans is their contents. One may want to try different sodas of
both diet and regular varieties to show this.

Absolute Zero Determination Demo

Investigation and Experimentation Ojectives

In this activity, the student will make and record observations, graph the measurement data, and make
use of the concepts of interpolation and extrapolation in the construction of a graph.

Brief description of demonstration

An Absolute Zero Apparatus is placed in various liquids at different temperatures. The temperatures of
each solution are known. The pressure is read from the pressure gauge on the apparatus. A graph is
made with Celsius temperature on the vertical axis and pressure on the horizontal axis. The plot is then
extrapolated to zero pressure. The extrapolated line will cross the temperature axis at absolute zero.

Apparatus and Materials

• Absolute zero apparatus (available from science supply companies for around $150)

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Figure 3.2

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108

3 - Pyrex or Kimex 2.0 liter beakers

Hotplate

Ice

Dry ice (you can find dry ice suppliers on the internet - dry ice can be stored in a Styrofoam cooler

but do not put the lid on tightly)

Ethanol - 600 mL

If you have a mercury thermometer that covers the range -100°C to +100°C, you can use it to

measure the temperatures of the baths.

Procedure

1. Fill one of the Pyrex beakers half full of tap water and place it on the hotplate to boil.

2. Fill the another Pyrex beaker half full of tap water and crushed ice.

3. Fill the third beaker about one-fourth full will broken pieces of dry ice and then add ethanol slowly
(lots of fog) until the beaker is about half full.

4. Place the bulb of the Absolute Zero Apparatus into the boiling water and leave it there until a
constant pressure is reached. Record the temperature of the bath (taken to be 100°C) and the
pressure on the gauge.

5. Place the bulb of the Absolute Zero Apparatus into the ice water and leave it there until a constant
pressure is reached. Record the temperature of the bath (taken to be 0°C) and the pressure on the
gauge.

6. Place the bulb of the device into the dry ice and alcohol slush and leave it there until a constant
pressure is reached. Record the temperature (known to be — 81°C) and the pressure on the gauge.

7. Plot a graph of the temperatures and pressures recorded. Make sure that the temperature axis on
your graph extends below — 280°C.

8. After the three points are plotted on the graph, lay a straight edge on the graph line and extend it
to the zero pressure line. You will get a graph similar to the one shown below.

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100

60
40

:o
o

-88

-40

o W _«.
jf

9 -80

g -100

a im

H -140
-1«0
-180

-200
-220
-240
-2«0
280
-300

1
9

*
f

0.1 0.2 0.3 0.4 0.5 0,6 0.7 0.8 1.0 1.2 1.3 1,4 1.5 1.6 1,7 l.S 1,9 2.0

Pressure, atm.

Hazards

Do not handle dry ice with bare hands. Pot holders or thermal protection gloves are necessary to handle
the boiling water beaker and the dry ice - alcohol slush beaker.

Disposal

Once the dry ice has all melted, the solutions can be poured down the sink.

Discussion

Pressure is caused by the collisions of gas particles with each other and the walls of their container. When
the temperature is lowered, the particles move more slowly, decreasing the frequency and strength of these
collisions. In turn, the pressure falls.

Absolute temperature can be defined as the temperature at which molecules cease to move. Therefore
absolute zero temperature corresponds to zero pressure.

Extending a graph beyond actual data points is called extrapolation ... a not-always acceptable procedure.

3.4 How Scientists Use Data

Lesson Objectives

Define the terms law, hypothesis, and theory.
Explain why scientists use models.

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110

Introduction

In the last section, we learned a little more about making quantitative and qualitative observations. A set
of observations about a particular phenomenon is called data. Scientist use many techniques to analyze
and interpret data. Data analysis produces organized data that is more conducive to seeing regularities
and drawing conclusions. Making tables and graphs of data are two of the most useful techniques in data
analysis.

Natural Laws are Statements of Repeated Data Patterns

Around the year 1800, Jacque Charles and other scientists were working with gases to, among other reasons,
improve the design of the hot air balloon. These scientists found, after many, many tests, that patterns
and regularities existed in the observations on gas behavior. If the temperature of the gas increased, the
volume of the gas increased. This is known as a natural law. A natural law is a relationship that exists
between variables in a group of data. Natural laws describe the patterns we see in large amounts of data.
These laws have withstood the test of time because they have been based on repeated observation with no
known exceptions.

Figure 3.3: Scuba Divers.

Around the same time as Charles was working with hot air balloons, another scientist, names J.W. Henry
was doing experiments trying to find a pattern between the pressure of a gas and the amount of the gas that
dissolved in water. Henry found that when one of these variables increased, the other variable increased in
the same proportion. Have you ever gone scuba diving (Figure 3.3)? Scuba Divers learn about a problem
known as "the Bends" when they are being trained. As scuba divers dive deeper, the increased pressure of
the breathing air causes more nitrogen to be dissolved in the diver's blood. Coming up too quickly from
a dive causes the pressure to decrease rapidly and therefore, the nitrogen to leave the blood quickly which
leads to "the bends." Henry's Law is called a natural law because it indicates a relationship (regularity)
between gas pressure and the amount of dissolved nitrogen.

Ill

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A Hypothesis is a Tentative Explanation

When scientists develop a description of the nature of matter to explain observations (including natural
laws), the first attempt at an explanation is often referred to as a hypothesis. A hypothesis is your educated
or best guess as to the nature of matter that causes those observations. The requirements for a hypothesis
are only that the hypothesis explains ALL the observations and that it is possible to make an observation
that will refute the hypothesis.

The hypothesis must be testable. The test of a hypothesis is called an experiment. If the results of the
experiment contradict the hypothesis, the hypothesis is rejected and a new hypothesis is formulated. The
results of the experiment are now included in the observations list and the new hypothesis must explain
this new observation as well as all the previous observations. If the result of the experiment supports the
hypothesis, more tests are still required. Hypotheses are not proven by testing . . . they are merely
supported or contradicted.

A Theory is an Explanation of a Law

As stated earlier in this section, a law describes a pattern of data that is observed with no known exception.
A theory is a possible explanation for a law. In science, theories can either be descriptive (qualitative) or
mathematical (quantitative), but because they explain the patterns described in the law, theory can be
used to predict future events. On a popular television show, mathematical theories are used to analyze
and describe behavior in order to predict future events. Hypotheses that have survived many supportive
tests are often called theories. Theories have a great deal more supportive testing behind them than do
hypotheses.

Let's put it together. The Law of Conservation of Mass that you learned earlier stated that matter cannot
be created nor destroyed. For example, in the reaction below, you will see that there are 28g + 6g = 34 g
of reactants (before the arrow) and 34 g of products (after the arrow).

N 2 ( g ) + 3 H 2 (g) -» 2 NH 3(g)

28g 6g 34g

Mass must remain constant from the start of a reaction to completion. This law was the result of many
quantitative experiments done by John Dalton and others in the early part of the 1800s. Dalton had
formulated many hypotheses surrounding his vision of how the elements worked together, how compounds
formed, and how chemical reactions would take place maintaining this mass from beginning to end. Even-
tually, in 1803, Dalton was able to propose the atomic theory which was an explanation for this law.

Sample Question: What distinguishes a law from a theory?

Solution: A law is an observation of nature; a theory is a possible explanation of the law.

Models Developed to Aid in Understanding

Scientists often use models when they need a way to communicate their understanding of what might be
very small (such as an atom or molecule) or very large (such as the universe). A model is another way to
express a theory.

If you were asked to determine the contents of a box that cannot be opened, you would do a variety of
experiments in order to develop an idea (or a model) of what the box contains. You would probably shake
the box, perhaps put magnets near it and/or determine its mass. When you completed your experiments,

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you would develop an idea of what is inside; that is, you would make a model of what is inside a box that
cannot be opened.

A good example of how a model is useful to scientists is the kinetic molecular theory. The theory can be
defined in statements, but it becomes much more easily understood if representations of the particles in
their three phases are drawn.

Another example is how models were used to explain the development of the atomic theory. As you will
learn in a later chapter, the idea of the concept of an atom changed over many years. In order to understand
each of the different theories of the atom according to the various scientists, models were drawn, and more
easily understood.

Lesson Summary

• A natural law is an observation, or a description of a large amount of reproducible data.

• A hypothesis is a early attempt at an explanation for data.

• A theory is used to explain a law or to explain a series of facts/events.

• Theories can use qualitative analogies or models to describe results.

Review Questions

1. Jack performed an experiment where he measured the masses of two different reactants and the
resulting product. His results are shown in the equation below. What law is Jack demonstrating in
his experiment?

10* 20# 30g

(a) law of constant composition

(b) law of combining volumes

(d) law of conservation of energy

(e) law of multiple proportions

2. Sugar dissolves in water. What kind of a statement is this?

(a) a hypothesis

(b) a law

(d) a rule

(e) all of the above

3. Draw a model to represent the difference between a solid, a liquid, and a gas. In your model, use
symbols to represent the molecules that are present in each state. The model should then show how
the molecules exist in each state.

Further Reading / Supplemental Links

• http : //en . wikipedia . org/wiki/Category : Chemistry_theories

• http : //en . wikipedia . org/wiki/Polytetraf luoroethylene

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Vocabulary

natural laws A description of the patterns observed in the large amounts of data.

hypothesis An educated guess as to what is going to happen in the experiment.

theory Used to explain a law or to explain a series of facts/events.

law of conservation of mass Matter cannot be created nor destroyed.

model A description, graphic, or 3-D representation of theory used to help enhance understanding.

scientific method The method of deriving the theories from hypotheses and laws through experimen-
tation and observation.

Image Sources

(1) http : //www . katrina . noaa . gov/satellite/satellite . html.

(2) Scuba Divers..

(3) Richard Parsons. . CCBYSA.

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Chapter 4
Atomic Theory

4.1 Early Development of a Theory

Lesson Objectives

• Give a short history of the Concept of the atom.

• State the Law of Definite Proportions.

• State the Law of Multiple Proportions.

• State Dalton's Atomic Theory, and explain its historical development.

Introduction

You learned earlier how all matter in the universe is made out of tiny building blocks called atoms. The
concept of the atom is accepted by all modern scientists, but when atoms were first proposed about 2500
years ago, ancient philosophers laughed at the idea. It has always been difficult to convince people of
the existence of things that are too small to see. There are many observations that are made on atoms,
however, that do not involve actually seeing the atom itself and science is about observing and devising
a theory to explain why those observations occur. We will spend some time considering the evidence
(observations) that convince scientists of the existence of atoms.

Democritus and the "Atom"

Before we discuss the experiments and evidence which have, over the years, convinced scientists that matter
is made up of atoms, it's only fair to give credit to the man who proposed "atoms" in the first place. 2,500
years ago, early Greek philosophers believed the entire universe was a single, huge, entity. In other words,
"everything was one." They believed that all objects, all matter, and all substance were connected as a
single, big, unchangeable "thing." Now you're probably disturbed by the word unchangeable. Certainly
you've seen the world around you change, and those early Greek philosophers must have too. Why, then,
would they think that the universe was unchangeable? Well, strange as it may sound to you today, back
then the generally accepted theory was that the world didn't change - it just looked like it did. In other
words, Greek philosophers believed that all change (and all motion) was an illusion. It was all in your
head! Compared to this crazy idea, the atom is looking pretty good, isn't it?

One of the first people to propose "atoms" was a man known as Democritus (Figure 4.1). Democritus

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Figure 4.1: Democritus was known as "The Laughing Philosopher." It's a good thing he liked to laugh,
because most other philosophers were laughing at his theories.

didn't like the idea that life was an illusion any more than you probably do. As an alternative, he suggested
that the world did change, and he explained this change by proposing atomos or atomon - tiny, indivisible,
solid objects making up all matter in the universe.

Democritus then reasoned that changes occur when the many atomos in an object were reconnected or
recombined in different ways. Democritus even extended his theory, suggesting that there were different
varieties of atomos with different shapes, sizes, and masses. He thought, however, that shape, size and
mass were the only properties differentiating the different types of atomos. According to Democritus, other
characteristics, like color and taste, did not reflect properties of the atomos themselves, but rather, resulted
from the different ways in which the atomos were combined and connected to one another (Figure 4.2).

Figure 4.2: Democritus believed that properties like color depended on how the atomos were connected to
each other, and not on the atomos themselves. Interestingly, Democritus was partially right - the green
emerald and the red ruby both contain atoms of aluminum, oxygen, and chromium. The emerald, however,
also contains silicon and beryllium atoms.

Even though the idea of the atomos seems much more reasonable than trying to explain experience as an
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illusion, the early Greek philosophers didn't like it, and they didn't for the following reason. If all matter
consists of tiny atomos that float around, bang into each other, and connect together in various ways,
these atomos must be floating in something. But what? Well, according to Democritus, the atoms floated
around in a void (empty space or "nothingness"). That does seem a bit strange, doesn't it? Certainly, if
you pound your fist on the desk in front of you, it doesn't feel like there's any empty space in it. What's
more, Greek philosophers thought that empty space was illogical. In order to exist, they argued, nothing
must be something, meaning nothing wasn't nothing, but that's a contradiction. Their arguments got
quite confusing, but the end result was that Greek philosophers dismissed Democritus' theory entirely.
Sadly, it took over two millennia before the theory of atomos (or "atoms," as they're known today) was
fully appreciated.

Greek Philosophers Didn't Experiment

Figure 4.3: Greek philosophers liked to think - they didn't, however, like to experiment all that much.

Early Greek philosophers disliked Democritus' theory of atomos because they believed a void, or complete
"nothingness," was illogical. To ancient thinkers, a theory that went against "logic" was far worse than a
theory that went against experience or observation. That's because Greek philosophers truly believed that,
above all else, our understanding of the world should rely on "logic." In fact, they argued that the world
couldn't be understood using our senses at all, because our senses could deceive us (these were, of course,
the same people who argued that all change in the world was an illusion). Therefore, instead of relying
on observation, Greek philosophers tried to understand the world using their minds and, more specifically,
the power of reason.

Unfortunately, when Greek philosophers applied reason to Democritus' theory, their arguments were in-
consistent. Democritus' void had to be "something" to exist, but at the same time it had to be "nothing"
to be a void. Greek philosophers were not willing to accept the idea that "nothing" could be "something"
- that seemed illogical.

Today, we call these contradictions (such as "nothing" is "something") paradoxes. Science is full of para-
doxes. Sometimes these paradoxes result when our scientific theories are wrong or incomplete, and some-
times they result because we make bad assumptions about what's "logical" and what isn't. In the case of

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the void, it turns out that "nothing" really can exist, so in a way, "nothing" is "something."

So how could the Greek philosophers have known that Democritus had a good idea with his theory of
"atomos?" It would have taken is some careful observation and a few simple experiments. Now you might
wonder why Greek philosophers didn't perform any experiments to actually test Democritus' theory. The
problem, of course, was that Greek philosophers didn't believe in experiments at all. Remember, Greek
philosophers didn't trust their senses, they only trusted the reasoning power of the mind (Figure 4.3).

Alchemists Experimented But Didn't Seek Explanation

Figure 4.4: Yet another alchemist searching for the philosopher's stone. Notice how the alchemists used a
lot of experimental techniques. It's too bad they were only interested in making gold!

As you learned in the last section, the early Greek philosophers tried to understand the nature of the world
through reason and logic, but not through experiment and observation. As a result, they had some very
interesting ideas, but they felt no need to justify their ideas based on life experiences. In a lot of ways, you
can think of the Greek philosophers as being "all thought and no action." It's truly amazing how much they
achieved using their minds, but because they never performed any experiments, they missed or rejected a
lot of discoveries that they could have made otherwise.

Some of the earliest experimental work was done by the alchemists (Figure 4.4). Remember that the
alchemists were extremely interested in discovering the "philosopher's stone", which could turn common
metals into gold. Of course, they also dabbled in medicines and cures, hoping to find "the elixir of life",
and other such miraculous potions. On the other hand, alchemists were not overly concerned with any
deep questions about the nature of the world. In contrast to the Greek philosophers, you can think of

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118

alchemists as being "all action and no thought." Alchemists experimented freely with everything that they
could find. In general, though, they didn't think too much about their results and what their results might
tell them about the world. Instead, they were only interested in whether or not they had made gold.

To be fair, there were some alchemists who tried to use results from past experiments to help suggest future
experiments. Nevertheless, alchemy always had very materialistic goals in mind - goals like producing gold
and living forever. Alchemists were not troubled by philosophical questions like "what is the universe made
of?" - they didn't really care unless they thought it would somehow help them find the "philosopher's stone"
or the "elixir of life."

What you've probably noticed by reading about the Greek philosophers and the alchemists is that the
history of science is ironic. Greek philosophers asked deep questions about the universe but didn't believe
in any of the experiments that might have led them to the answers. In contrast, alchemists believed in
experimentation but weren't interested in what the experiments might tell them in terms of the nature
of the world. Unbelievably, it took over 2000 years to put the questions asked by the Greek philosophers
together with the experimental tools developed by the alchemists. The result was significant progress in
our understanding of nature and the universe, and that's what we'll learn about next.

Dalton's Atomic Theory

Let's begin our discussion of Dalton's atomic theory by considering a simple, but important experiment that
suggested matter might be made up of atoms. In the late 1700's and early 1800's, scientists began noticing
that when certain substances, like hydrogen and oxygen, were combined to produce a new substance, like
water, the reactants (hydrogen and oxygen) always reacted in the same proportions by mass. In other
words, if 1 gram of hydrogen reacted with 8 grams of oxygen, then 2 grams of hydrogen would react
with 16 grams of oxygen, and 3 grams of hydrogen would react with 24 grams of oxygen. Strangely, the
observation that hydrogen and oxygen always reacted in the "same proportions by mass" wasn't special.
In fact, it turned out that the reactants in every chemical reaction reacted in the same proportions by
mass. Take, for example, nitrogen and hydrogen, which react to produce ammonia (a chemical you've
probably used to clean your house). 1 gram of hydrogen will react with 4.7 grams of nitrogen, and 2 grams
of hydrogen will react with 9.4 grams of nitrogen. Can you guess how much nitrogen would react with
3 grams of hydrogen?

Scientists studied reaction after reaction, but every time the result was the same. The reactants always
reacted in the "same proportions by mass" or in what we call "definite proportions." As a result, scientists
proposed the Law of Definite Proportions (Figure 4.5). This law states that:

In a given type of chemical substance, the elements are always combined in the same pro-
portions by mass.

Earlier, you learned that an "element" is a grouping in which there is only one type of atom - of course,
when the Law of Definite Proportions was first discovered, scientists didn't know about atoms or elements,
so the law was stated slightly differently. We'll stick with this modern version, though, since it's easiest to
understand.

The Law of Definite Proportions applies when elements are reacted together to form the same product.
Therefore, while the Law of Definite Proportions can be used to compare two experiments in which hydrogen
and oxygen react to form water, the Law of Definite Proportions can not be used to compare one experiment
in which hydrogen and oxygen react to form water, and another experiment in which hydrogen and oxygen
react to form hydrogen peroxide (peroxide is another material that can be made from hydrogen and oxygen) .

While scientists around the turn of the 18 century weren't making a lot of peroxide, a man named John
Dalton was experimenting with several reactions in which the reactant elements formed more than one type
of product, depending on the experimental conditions he used (Figure 4.6). One common reaction that he

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A
A

Figure 4.5: If of reacts with of , then by the Law of Definite Proportions, of must react with of . If of
reacts with of , then by the Law of Conservation of Mass, they must produce of . Similarly, when of react
with of , they must produce of .

Figure 4.6: John Dalton was a thinker, but he was also an experimenter.

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studied was the reaction between carbon and oxygen. When carbon and oxygen react, they produce two
different substances - we'ii caff these substances "A" and "B." It turned out that, given the same amount of
carbon, forming B always required exactly twice as much oxygen as forming A. In other words, if you can
make A with 3 grams of carbon and 4 grams of oxygen, B can be made with the same 3 grams of carbon,
but with 8 grams oxygen. Dalton asked himself - why does B require 2 times as much oxygen as A? Why
not 1.21 times as much oxygen, or 0.95 times as much oxygen? Why a whole number like 2?

The situation became even stranger when Dalton tried similar experiments with different substances. For
example, when he reacted nitrogen and oxygen, Dalton discovered that he could make three different
substances - we'll call them "C," "D," and "£." As it turned out, for the same amount of nitrogen, D
always required twice as much oxygen as C. Similarly, E always required exactly four times as much
oxygen as C. Once again, Dalton noticed that small whole numbers (2 and 4) seemed to be the rule.

Dalton used his experimental results to propose The Law of Multiple Proportions:

When two elements react to form more than one substance, the different masses of one
element (like oxygen) that are combined with the same mass of the other element (like
nitrogen) are in a ratio of small whole numbers.

Dalton thought about his Law of Multiple Proportions and tried to find some theory that would explain it.
Dalton also knew about the Law of Definite Proportions and the Law of Conservation of Mass (Remember
that the Law of Conservation of Mass states that mass is neither created nor destroyed), so what he really
wanted was a theory that would explain all three of these laws using a simple, plausible model. One way to
explain the relationships that Dalton and others had observed was to suggest that materials like nitrogen,
carbon and oxygen were composed of small, indivisible quantities which Dalton called "atoms" (in reference
to Democritus' original idea). Dalton used this idea to generate what is now known as Dalton' s Atomic
Theory*

Dalton's Atomic Theory

1. Matter is made of tiny particles called atoms.

2. Atoms are indivisible. During a chemical reaction, atoms are rearranged, but they do not break
apart, nor are they created or destroyed.

3. All atoms of a given element are identical in mass and other properties.

4. The atoms of different elements differ in mass and other properties.

5. Atoms of one element can combine with atoms of another element to form "compounds" - new,
complex particles. In a given compound, however, the different types of atoms are always present in
the same relative numbers.

*Some people think that Dalton developed his Atomic Theory before stating the Law of Multiple Propor-
tions, while others argue that the Law of Multiple Proportions, though not formally stated, was actually
discovered first. In reality, Dalton was probably contemplating both concepts at the same time, although
it is hard to tell from his laboratory notes.

Lesson Summary

• 2,500 years ago, Democritus suggested that all matter in the universe was made up of tiny, indivisible,
solid objects he called "atomos."

• Democritus believed that there were different types of "atomos" which differed in shape, size, and
mass.

• Other Greek philosophers disliked Democritus' "atomos" theory because they felt it was illogical.
Since they didn't believe in experiments, though, they had no way to test the "atomos" theory.

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• Alchemists experimented and developed experimental techniques, but they were more interested in
making gold than they were in understanding the nature of matter and the universe.

• The Law of Definite Proportions states that in a given chemical substance, the elements are always
combined in the same proportions by mass.

• The Law of Multiple Proportions states that when two elements react to form more than one sub-
stance, the different masses of one element that are combined with the same mass of the other element
are in a ratio of small whole numbers.

• Dalton used the Law of Definite Proportions, the Law of Multiple Proportions, and The Law of
Conservation of Mass to propose his Atomic Theory.

• Dalton's Atomic Theory states:

1. Matter is made of tiny particles called atoms.

2. Atoms are indivisible. During a chemical reaction, atoms are rearranged, but they do not break
apart, nor are they created or destroyed.

3. All atoms of a given element are identical in mass and other properties.

4. The atoms of different elements differ in mass and other properties.

5. Atoms of one element can combine with atoms of another element to form "compounds" - new
complex particles. In a given compound, however, the different types of atoms are always present in
the same relative numbers.

Review Questions

1. It turns out that a few of the ideas in Dalton's Atomic Theory aren't entirely correct. Are inaccurate
theories an indication that science is a waste of time?

2. Suppose you are trying to decide whether to wear a sweater or a T-shirt. To make your decision, you
phone two friends. The first friend says, "Wear a sweater, because I've already been outside today,
and it's cold." The second friend, however, says, "Wear a T-shirt. It isn't logical to wear a sweater in
July." Would you decide to go with your first friend, and wear a sweater, or with your second friend,
and wear a T-shirt? Why?

3. Decide whether each of the following statements is true or false.

(a) Democritus believed that all matter was made of "atomos."

(b) Democritus also believed that there was only one kind of "atomos."

(d) If the early Greek philosophers hadn't been so interested in making gold, they probably would
have liked the idea of the "atomos."

4. Match the person, or group of people, with their role in the development of chemistry.

(a) Early Greek philosophers - a. suggested that all matter was made up of tiny, indivisible objects

(b) alchemists - b. tried to apply logic to the world around them

(d) Democritus - d. were primarily concerned with finding ways to turn common metals into gold

5. Early Greek philosophers felt that Democritus "atomos" theory was illogical because:

(a) no matter how hard they tried, they could never break matter into smaller pieces.

(b) it didn't help them to make gold.

(d) empty space is illogical because it implies that nothing is actually something.

6. Which Law explains the following observation: carbon monoxide can be formed by reacting 12 grams
of carbon with 16 grams of oxygen? To form carbon dioxide, however, 12 grams of carbon must react
with 32 grams of oxygen.

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7. Which Law explains the following observation: carbon monoxide can be formed by reacting 12 grams
of carbon with 16 grams of oxygen? It can also be formed by reacting 24 grams of carbon with
32 grams of oxygen.

8. Which Law explains the following observation: 28 grams of carbon monoxide are formed when
12 grams of carbon reacts with 16 grams of oxygen?

9. Which Law explains the following observations: when 12 grams of carbon react with 4 grams of
hydrogen, they produce methane, and there is no carbon or hydrogen left over at the end of the
reaction? If, however, 11 grams of carbon react with 4 grams of hydrogen, there is hydrogen left over
at the end of the reaction.

10. Which of the following is not part of Dalton's Atomic Theory?

(a) matter is made of tiny particles called atoms.

(b) during a chemical reaction, atoms are rearranged.

(d) all atoms of a specific element are the same.

Calculations

11. Consider the following data: 3.6 grams of boron react with 1.0 grams of hydrogen to give 4.6 grams
of BH3. How many grams of boron would react with 2.0 grams of hydrogen?

12. Consider the following data: 12 grams of carbon and 4 grams of hydrogen react to give 16 grams of
"compound A." 24 grams of carbon and 6 grams of hydrogen react to give 30 grams of "compound
5." Are compound A and compound B the same? Why or why not?

Vocabulary

atomos (atomon) Democritus' word for the tiny, indivisible, solid objects that he believed made up all

matter in the universe.
void Another word for empty space.

paradox Two statements that seem to be true, but contradict each other.

law of definite proportions In a given chemical substance, the elements are always combined in the
same proportions by mass.

law of multiple proportions When two elements react to form more than one substance, the different
masses of one element that are combined with the same mass of the other element are in a ratio of
small whole numbers.

4.2 Further Understanding of the Atom

Lesson Objectives

• Explain the experiment that led to Thomson's discovery of the electron.

• Describe Thomson's "plum pudding" model of the atom.

• Describe Rutherford's Gold Foil experiment, and explain how this experiment proved that the "plum
pudding" model of the atom was incorrect.

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Introduction

In the last lesson, you learned about the atom, and the early experiments that led to the development of
Dalton's Atomic Theory. But Dalton's Atomic Theory isn't the end of the story. Do you remember the
scientific method introduced in early on in the book? Chemists using the scientific method make careful
observations and measurements and then use these measurements to propose theories. That's exactly what
Dalton did. Dalton used the following observations:

1. Mass is neither created nor destroyed during a chemical reaction.

2. Elements always combine in the same proportions by mass when they form a given compound.

3. When elements form more than one compound, the different masses of one element that are combined
with the same mass of the other element are in a ratio of small whole numbers.

With these observations (which came from careful measurement), Dalton proposed his Atomic Theory - a
model which suggested how the underlying structure of matter might lead to the three observations above.
The scientific method, though, doesn't stop once a theory has been proposed. Instead, the theory should
suggest new experiments that can be performed to test whether or not the original theory is accurate and
complete.

Dalton's Atomic Theory held up well to a lot of the different chemical experiments that scientists performed
to test it. In fact, for almost 100 years, it seemed as if Dalton's Atomic Theory was the whole truth.
However, in 1897, a scientist named J. J. Thompson conducted some research which suggested that Atomic
Theory wasn't the entire story. As it turns out, Dalton had a lot right. He was right in saying matter is
made up of atoms; he was right in saying there are different kinds of atoms with different mass and other
properties; he was "almost" right in saying atoms of a given element are identical; he was right in saying
during a chemical reaction, atoms are merely rearranged; he was right in saying a given compound always
has atoms present in the same relative numbers. But he was WRONG in saying atoms were indivisible
or indestructible. As it turns out, atoms are divisible. In fact, atoms are composed of smaller subatomic
particles. We'll talk about the discoveries of these subatomic particles next.

Thomson Discovered Electrons Were Part of the Atom

In the mid- 1800s, scientists were beginning to realize that the study of chemistry and the study of electricity
were actually related. First, a man named Michael Faraday showed how passing electricity through mixtures
of different chemicals could cause chemical reactions. Shortly after that, scientists found that by forcing
electricity through a tube filled with gas, the electricity made the gas glow! Scientists didn't, however,
understand the relationship between chemicals and electricity until a British physicist named J.J. Thomson
(Figure 4.7) began experimenting with what is known as a cathode ray tube.

Figure ?? shows a basic diagram of a cathode ray tube like the one J. J. Thomson would have used. A
cathode ray tube is a small glass tube with a cathode (a negatively charged metal plate) and an anode (a
positively charged metal plate) at opposite ends. By separating the cathode and anode by a short distance,
the cathode ray tube can generate what are known as "cathode rays" - rays of electricity that flowed from
the cathode to the anode, J. J. Thomson wanted to know what cathode rays were, where cathode rays
came from and whether cathode rays had any mass or charge. The techniques that J. J. Thomson used
to answer these questions were very clever and earned him a Nobel Prize in physics. First, by cutting a
small hole in the anode J. J. Thomson found that he could get some of the cathode rays to flow through
the hole in the anode and into the other end of the glass cathode ray tube. Next, J. J. Thomson figured
out that if he painted a substance known as "phosphor" onto the far end of the cathode ray tube, he could
see exactly where the cathode rays hit because the cathode rays made the phosphor glow.

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Figure 4.7: A portrait of J. J. Thomson.

A cathode-ray tube. (Source: Sharon Bewick. CC-BY-SA)

electricity passes from the
cathode to the anode

a small bit of electricity
passes through the hole
in the anode

cathode

phophor coating

J. J. Thomson must have suspected that cathode rays were charged, because his next step was to place a
positively charged metal plate on one side of the cathode ray tube and a negatively charged metal plate
on the other side of the cathode ray tube, as shown in (Figure ??). The metal plates didn't actually
touch the cathode ray tube, but they were close enough that a remarkable thing happened! The flow of
the cathode rays passing through the hole in the anode was bent upwards towards the positive metal plate
and away from the negative metal plate.

A cathode ray was attracted to the positively charged metal plate placed above the tube, and repelled
from negatively charged metal plate placed below the tube. (Source: Sharon Bewick. CC-BY-SA)

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positively charged

metal plate

negatively charged
metal plate

In other words, instead of the phosphor glowing directly across from the hole in the anode (as in Figure
??), the phosphor now glowed at a spot quite a bit higher in the tube (as in Figure ??).

J. J. Thomson thought about his results for a long time. It was almost as if the cathode rays were attracted
to the positively charged metal plate above the cathode ray tube, and repelled from the negatively charged
metal plate below the cathode ray tube. J. J. Thomson knew that charged objects are attracted to and
repelled from other charged objects according to the rule: opposites attract, likes repel. This means that
a positive charge is attracted to a negative charge, but repelled from another positive charge. Similarly,
a negative charge is attracted to a positive charge, but repelled from another negative charge. Using the
"opposites attract, likes repel" rule, J. J. Thomson argued that if the cathode rays were attracted to the
positively charged metal plate and repelled from the negatively charged metal plate, they themselves must
have a negative charge!

J. J. Thomson then did some rather complex experiments with magnets, and used his results to prove that
cathode rays were not only negatively charged, but also had mass. Remember that anything with mass is
part of what we call matter. In other words, these cathode rays must be the result of negatively charged
"matter" flowing from the cathode to the anode. But there was a problem. According to J. J. Thomson's
measurements, either these cathode rays had a ridiculously high charge, or else had very, very little mass -
much less mass than the smallest known atom. How was this possible? How could the matter making up
cathode rays be smaller than an atom if atoms were indivisible? J. J. Thomson made a radical proposal:
maybe atoms are divisible. J. J. Thomson suggested that the small, negatively charged particles making
up the cathode ray were actually pieces of atoms. He called these pieces "corpuscles," although today we
know them as "electrons." Thanks to his clever experiments and careful reasoning J. J. Thomson is credited
with the discovery of the electron.

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126

Protons Were Thought to Exist but Discovered Much Later

In the last section, we learned that atoms are, in fact, divisible, and that one of the subatomic particles
making up an atom is a small, negatively charged entity called an "electron." Now imagine what would
happen if atoms were made entirely of electrons. First of all, electrons are very, very small; in fact, electrons
are about 2000 times smaller than the smallest known atom, so every atom would have to contain a whole
lot of electrons. But there's another, even bigger problem: electrons are negatively charged. Therefore, if
atoms were made entirely out of electrons, atoms would be negatively charged themselves... and that would
mean all matter was negatively charged as well.

Of course, matter isn't negatively charged. In fact, most matter is what we call neutral - it has no charge
at all. If matter is composed of atoms, and atoms are composed of negative electrons, how can matter
be neutral? The only possible explanation is that atoms consist of more than just electrons. Atoms must
also contain some type of positively charged material which balances the negative charge on the electrons.
Negative and positive charges of equal size cancel each other out, just like negative and positive numbers
of equal size. What do you get if you add +1 and -1? You get 0, or nothing. That's true of numbers,
and that's also true of charges. If an atom contains an electron with a -1 charge, but also some form of
material with a +1 charge, overall the atom must have a (+1) + (-1) = charge - in other words, the
atom must be neutral, or have no charge at all.

Based on the fact that atoms are neutral, and based on J. J. Thomson's discovery that atoms contain
negative subatomic particles called "electrons," scientists assumed that atoms must also contain a positive
substance. It turned out that this positive substance was another kind of subatomic particle, known
as the "proton." Although scientists knew that atoms had to contain positive material, protons weren't
actually discovered, or understood, until quite a bit later.

Thomson's Model of the Atom

Figure 4.8: A plum pudding and Thomson's "plum-pudding" model for the atom. Notice how the "plums"
are the negatively charged electrons, while the positive charge is spread throughout the entire pudding
batter.

When Thomson discovered the negative electron, he realized that atoms had to contain positive material
as well - otherwise they wouldn't be neutral overall. As a result, Thomson formulated what's known as
the "plum pudding" model for the atom. According to the "plum pudding" model, the negative electrons
were like pieces of fruit and the positive material was like the batter or the pudding. This made a lot of
sense given Thomson's experiments and observations. Thomson had been able to isolate electrons using

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a cathode ray tube; however he had never managed to isolate positive particles. As a result, Thomson
theorized that the positive material in the atom must form something like the "batter" in a plum pudding,
while the negative electrons must be scattered through this "batter." (If you've never seen or tasted a plum
pudding, you can think of a chocolate chip cookie instead. In that case, the positive material in the atom
would be the "batter" in the chocolate chip cookie, while the negative electrons would be scattered through
the batter like chocolate chips.)

Figure 4.8 shows a "plum pudding" and a "plum pudding" model for the atom. Notice how easy it would
be to pick the pieces of fruit out of a plum pudding. On the other hand, it would be a lot harder to
pick the batter out of the plum pudding, because the batter is everywhere. If an atom were similar to a
plum pudding in which the electrons are scattered throughout the "batter" of positive material, then you'd
expect it would be easy to pick out the electrons, but a lot harder to pick out the positive material.

Everything about Thomson's experiments suggested the "plum pudding" model was correct - but accord-
ing to the scientific method, any new theory or model should be tested by further experimentation and
observation. In the case of the "plum pudding" model, it would take a man named Ernest Rutherford to
prove it wrong. Rutherford and his experiments will be the topic of the next section.

Rutherford's Model of the Atom

Disproving Thomson's "plum pudding" model began with the discovery that an element known as uranium
emitted positively charged particles called alpha particles as it underwent radioactive decay. Radioactive
decay occurs when one element decomposes into another element. It only happens with a few very unstable
elements. This involves some difficult concepts so, for now, just accept the fact that uranium decays and
emits alpha particles in the process. Alpha particles themselves didn't prove anything about the structure
of the atom. In fact, a man named Ernest Rutherford proved that alpha particles were nothing more than
helium atoms that had lost their electrons (Figure ??). Think about why an atom that has lost electrons
will have a positive charge. Alpha particles could, however, be used to conduct some very interesting
experiments.

Ernest Rutherford was fascinated by all aspects of alpha particles. For the most part, though, he seemed
to view alpha particles as tiny bullets that he could use to fire at all kinds of different materials. One
experiment in particular, however, surprised Rutherford, and everyone else. Rutherford found that when he

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fired alpha particles at a very thin piece of gold foil, an interesting thing happened (Figure ??). Almost
all of the alpha particles went straight through the foil as if they'd hit nothing at all. Every so often,
though, one of the alpha particles would be deflected slightly as if it had bounced off of something hard.
Even less often, Rutherford observed alpha particles bouncing straight back at the "gun" from which they
had been fired! It was as if these alpha particles had hit a wall "head-on" and had ricocheted right back
in the direction that they had come from.

Rutherford thought that these experimental results were rather odd. Rutherford described firing alpha
particles at gold foil like shooting a high-powered rifle at tissue paper. Would you ever expect the bullets
to hit the tissue paper and bounce back at you? Of course not! The bullets would break through the tissue
paper and keep on going, almost as if they'd hit nothing at all. That's what Rutherford had expected
would happen when he fired alpha particles at the gold foil. Therefore, the fact that most alpha particles
passed through didn't shock him. On the other hand, how could he explain the alpha particles that got
deflected? Even worse, how could he explain the alpha particles that bounced right back as if they'd hit a
wall?

Rutherford decided that the only way to explain his results was to assume that the positive matter forming
the gold atoms was not, in fact, distributed like the batter in plum pudding, but rather, was concentrated
in one spot, forming a small positively charged particle somewhere in the center of the gold atom. We
now call this clump of positively charged mass the nucleus. According to Rutherford, the presence of a
nucleus explained his experiments, because it implied that most alpha particles passed through the gold
foil without hitting anything at all. Once in a while, though, the alpha particles would actually collide with
a gold nucleus, causing the alpha particles to be deflected, or even to bounce right back in the direction
they came from.

a particles

nucleus go

deflected
Ct, particles

non-deflected
a particles

gold atom

A short history of the changes in our model of the atom, an image of the plum pudding model, and an
animation of Rutherford's experiment can be viewed at Plum Pudding and Rutherford Page (http : //www .
newcastle-schools . org . uk/nsn/chemistry/Rad.ioactivity/Plub /„20Pudding /o20and. / 20Rutherf ord°/ 20Page .
htm).

129

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Rutherford Suggested Electrons "Orbited"

While Rutherford's discovery of the positively charged atomic nucleus offered insight into the structure of
the atom, it also led to some questions. According to the "plum pudding" model, electrons were like plums
embedded in the positive "batter" of the atom. Rutherford's model, though, suggested that the positive
charge wasn't distributed like batter, but rather, was concentrated into a tiny particle at the center of the
atom, while most of the rest of the atom was empty space. What did that mean for the electrons? If
they weren't embedded in the positive material, exactly what were they doing? And how were they held in
the atom? Rutherford suggested that the electrons might be circling or "orbiting" the positively charged
nucleus as some type of negatively charged cloud, but at the time, there wasn't much evidence to suggest
exactly how the electrons were held in the atom.

Despite the problems and questions associated with Rutherford's experiments, his work with alpha particles
definitely seemed to point to the existence of an atomic "nucleus." Between J. J. Thomson, who discovered
the electron, and Rutherford, who suggested that the positive charges in an atom were concentrated at the
atom's center, the 1890s and early 1900s saw huge steps in understanding the atom at the "subatomic" (or
smaller than the size of an atom) level. Although there was still some uncertainty with respect to exactly
how subatomic particles were organized in the atom, it was becoming more and more obvious that atoms
were indeed divisible. Moreover, it was clear that the pieces an atom could be separated into negatively
charged electrons and a nucleus containing positive charges. In the next lesson, we'll look more carefully
at the structure of the nucleus, and we'll learn that while the atom is made up of positive and negative
particles, it also contains neutral particles that neither Thomson, nor Rutherford, were able to detect with
their experiments.

Lesson Summary

• Dalton's Atomic Theory wasn't entirely correct. It turns out that atoms can be divided into smaller
subatomic particles.

• A cathode ray tube is a small glass tube with a cathode and an anode at one end. Cathode rays flow
from the cathode to the anode.

• When cathode rays hit a material known as "phosphor" they cause the phosphor to glow. J. J.
Thomson used this phenomenon to reveal the path taken by a cathode ray in a cathode ray tube.

• J. J. Thomson found that the path taken by the cathode ray could be bent towards a positive metal
plate, and away from a negative metal plate. As a result, he reasoned that the particles in the cathode
ray were negative.

• Further experiments with magnets proved that the particles in the cathode ray also had mass. Thom-
son's measurements indicated, however, that the particles were much smaller than atoms.

• J. J. Thomson suggested that these small, negatively charged particles were actually subatomic
particles. We now call them "electrons."

• Since atoms are neutral, atoms that contain negatively charged electrons must also contain positively
charged material.

• According to Thomson's "plum pudding" model, the negatively charged electrons in an atom are like
the pieces of fruit in a plum pudding, while the positively charged material is like the batter.

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• When Ernest Rutherford fired alpha particles at a thin gold foil, most alpha particles went straight
through; however, a few were scattered at different angles, and some even bounced straight back.

• In order to explain the results of his Gold Foil experiment, Rutherford suggested that the positive
matter in the gold atoms was concentrated at the center of the gold atom in what we now call the
nucleus of the atom.

• Rutherford's model of the atom didn't explain where electrons were located in an atom.

Review Questions

Concepts

1. Decide whether each of the following statements is true or false.

(a) Cathode rays are positively charged.

(b) Cathode rays are rays of light, and thus they have no mass.

(d) J.J. Thomson is credited with the discovery of the electron.

(e) Phosphor is a material that glows when struck by cathode rays.

2. Match each observation with the correct conclusion.

(a) Cathode rays are attracted to a positively charged metal plate. - i. Cathode rays are positively
charged. - ii. Cathode rays are negatively charged. - iii. Cathode rays have no charge.

(b) Electrons have a negative charge. - i. atoms must be negatively charged. - ii. atoms must be
positively charged. - iii. atoms must also contain positive subatomic material.

(c) Alpha particles fired at a thin gold foil are occasionally scattered back in the direction that
they came from - i. the positive material in an atom is spread throughout like the "batter" in
pudding - ii. atoms contain neutrons - iii. the positive charge in an atom is concentrated in a
small area at the center of the atom.

3. Alpha particles are:

(a) Helium atoms that have extra electrons.

(b) Hydrogen atoms that have extra electrons.

(d) Electrons.

(e) Helium atoms that have lost their electrons.

(f) Neutral helium atoms.

4. What is the name given to the tiny clump of positive material at the center of an atom?

5. Choose the correct statement.

(a) Ernest Rutherford discovered the atomic nucleus by performing experiments with aluminum
foil.

(b) Ernest Rutherford discovered the atomic nucleus using a cathode ray tube.

(d) Ernest Rutherford proved that the "plum pudding model" was incorrect.

(e) Ernest Rutherford experimented by firing cathode rays at gold foil.

6. Answer the following questions:

(a) Will the charges +2 and -2 cancel each other out?

(b) Will the charges +2 and -1 cancel each other out?

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(d) Will the charges -1 and +3 cancel each other out?

(e) Will the charges +9 and -9 cancel each other out?

7. Electrons are negatively charged metals plates and positively charged metal

plates?

8. What was J. J. Thomson's name for electrons?

9. A "sodium cation" is a sodium atom that has lost one of its electrons. Would the charge on a sodium
cation be positive, negative or neutral? Would sodium cations be attracted to a negative metal plate,
or a positive metal plate? Would electrons be attracted to or repelled from sodium cations?

10. Suppose you have a cathode ray tube coated with phosphor so that you can see where on the tube the
cathode ray hits by looking for the glowing spot. What will happen to the position of this glowing
spot if:

(a) a negatively charged metal plate is placed above the cathode ray tube

(b) a negatively charged metal plate is placed to the right of the cathode ray tube

(d) a negatively charged metal plate is placed above the cathode ray tube, and a positively charged
metal plate is placed to the left of the cathode ray tube

(e) a positively charged metal plate is placed below the cathode ray tube, and a positively charged
metal plate is also placed to the left of the cathode ray tube.

Vocabulary

subatomic particles Particles that are smaller than the atom. The three main subatomic particles are
electrons, protons and neutrons.

cathode rays rays of electricity that flow from the cathode to the anode. J.J. Thomson proved that
these rays were actually negatively charged subatomic particles (or electrons).

cathode A negatively charged metal plate.

anode A positively charged metal plate.

cathode ray tube A glass tube with a cathode and anode, separated by some distance, at one end.
Cathode ray tubes generate cathode rays.

phosphor A chemical that glows when it is hit by a cathode ray.

plum pudding model A model of the atom which suggested that the negative electrons were like plums
scattered through the positive material (which formed the batter).

alpha (a) particles Helium atoms that have lost their electrons. They are produced by uranium as it
decays.

nucleus The small central core of the atom where most of the mass of the atom (and all of the atoms
positive charge) is located.

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4.3 Atomic Terminology

Lesson Objectives

• Describe the properties of electrons, protons, and neutrons.

• Define and use an atom's atomic number (Z) and mass number (A).

• Define an isotope, and explain how isotopes affect an atom's mass, and an element's atomic mass.

Introduction

Dalton's Atomic Theory explained a lot about matter, chemicals, and chemical reactions. Nevertheless, it
wasn't entirely accurate, because contrary to what Dalton believed, atoms can, in fact, be broken apart
into smaller subunits or subatomic particles. One type of subatomic particle found in an atom is the
negatively charged electron. Since atoms are neutral, though, they also have to contain positive material.
At first, scientists weren't sure exactly what this positive material was, or how it existed in the atom.
Thomson thought it was distributed throughout the atom like batter in a plum pudding. Rutherford,
however, showed that this was not the case. In his Gold Foil experiment, Rutherford proved that the
positive substance in an atom was concentrated in a small area at the center of the atom, leaving most the
rest of the atom as empty space (possibly with a few electrons, or an "electron cloud").

Both Thomson's experiments and Rutherford's experiments answered a lot of questions, but they also
raised a lot of questions, and scientists wanted to know more. How were the electrons connected to the
rest of the atom? What was the positive material at the center of the atom like? Was it one giant clump
of positive mass, or could it be divided into smaller parts as well? In this lesson, we'll look at the atom a
little more closely.

Electrons, Protons, and Neutrons

The atom is composed of three different kinds of subatomic particles. First, there are the electrons, which
we've already talked about, and which J. J. Thomson discovered. Electrons have a negative charge. As
a result they are attracted to positive objects, and repelled from negative objects, which means that they
actually repel each other (Figure 4.9).

Neutral objects neither attract or repel.

Objects with the same charge repel.

Figure 4.9: Electrons repel each other because they are both negatively charged.

Still, most atoms have more than one electron. That's because atoms are big enough to hold many electrons
without those electrons ever colliding with each other. As you might expect, the bigger the atom, the more
electrons it contains.

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Protons are another type of subatomic particle found in atoms. Protons have a positive charge. As a result
they are attracted to negative objects, and repelled from positive objects. Again, this means that protons
repel each other (Figure 4.10). Unlike electrons, however, which manage to stake out a 'territory' and
'defend' it from other electrons, protons are bound together by what are termed strong nuclear forces.
Therefore, even though they repel each other, protons are forced to group together into one big clump.
This clump of protons helps to form the nucleus of the atom. Remember, the nucleus of the atom is the
mass of positive charge at the atom's center.

Neutral objects neither attract or repel.

Objects with the same charge repel.

Figure 4.10: Protons repel each other because they are both positively charged. Despite this repulsion,
protons are bound together in the atomic nucleus as a result of the strong nuclear force.

Electrons were the first subatomic particles discovered and protons were the second. There's a third kind of
subatomic particle, though, known as a neutron, which wasn't discovered until much later. As you might
have already guessed from its name, the neutron is neutral. In other words, it has no charge whatsoever,
and is therefore neither attracted to nor repelled from other objects. That's part of the reason why the
neutron wasn't discovered until long after people knew about electrons and protons - because it has no
charge, it's really hard to detect. Neutrons are in every atom (with one exception), and they're bound
together with other neutrons and protons in the atomic nucleus. Again, the binding forces that help to
keep neutrons fastened into the nucleus are known as strong nuclear forces.

Before we move on, we must discuss how the different types of subatomic particles interact with each other.
When it comes to neutrons, the answer is obvious. Since neutrons are neither attracted to, nor repelled
from objects, they don't really interact with protons or electrons (beyond being bound into the nucleus
with the protons). Protons and electrons, however, do interact. Using what you know about protons and
electrons, what do you think will happen when an electron approaches a proton - will the two subatomic
particles be attracted to each other, or repelled from each other? Here's a hint: "opposites attract, likes
repel." Electrons and protons have opposite charges (one negative, the other positive), so you'd expect them
to be attracted to each other and that's exactly what happens (Figure 4.11).

Viewers journey inside the atom to appreciate its architectural beauty and grasp how atomic structure
determines chemical behavior. Video on Demand - The World of Chemistry - The Atom (http : //www .
learner . org/vod/vod_window . html?pid=798)

Relative Mass and Charge

Even though electrons, protons, and neutrons are all types of subatomic particles, they are not all the same
size. When you compare the masses of electrons, protons and neutrons, what you find is that electrons
have an extremely small mass, compared to either protons or neutrons. On the other hand, the masses of
protons and neutrons are fairly similar, although technically, the mass of a neutron is slightly larger than
the mass of a proton. Because protons and neutrons are so much more massive than electrons, almost all
of the atomic mass in any atom comes from the nucleus, which contains all of the neutrons and protons.

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134

Neutral objects neither attract or repel.

Oppositely charged object? attract.

Figure 4.11: Protons and electrons are attracted to each other because they have opposite charges. Protons
are positively charged, while electrons are negatively charged.

Table 4.1: Masses of the Different Subatomic Particles

Mass in Grams (g)

Mass in Atomic Mass Units
(amu)

Electron

Proton

Neutron

9.109383 x 1(T 28
1.6726217 xl0~ 24
1.6749273 x 10~ 24

5.485799095 x 10~ 4

1.0072764669

1.0086649156

Table 4.1 gives the masses of electrons, protons, and neutrons. The second column shows the masses of the
three subatomic particles in grams (which is related to the SI unit kilograms according to the relationship
1 kg = 1000 g). The third column, however, shows the masses of the three subatomic particles in "atomic
mass units". Atomic mass units (amu) are useful, because, as you can see, the mass of a proton and the
mass of a neutron are almost exactly 1.0 in this unit system. We'll discuss atomic mass units in a later
section.

Unfortunately, the numbers in Table 4.1 probably don't give you a very good sense of just how big protons
and neutrons are compared to electrons, so here's a comparison that might help. If an electron were the
size of a penny, then a proton (or a neutron) would be about the size of a large bowling ball (Figure 4.12).

Figure 4.12: Electrons are much smaller than protons or neutrons. How much smaller? If an electron was
the size of a penny, a proton or a neutron would have the mass of a large bowling ball!

Obviously, if you were told to lift a box containing ^several bowling balls and several pennies,^8vFwou'rcm r 'f
really care about the pennies, because they wouldn't change the weight of the box all that much. What
you'd want to know, though, would be the number of bowling balls in the box. That's exactly what
harmens when scientists trv to fipmre nut the masses of atoms. Thev Hnn't reallv care hnw mamv electrons

"elementary charge units"* or "elementary charges." Elementary charge units (e) are appealing, because
the charge on a proton and the charge on an electron are exactly 1.0 in this unit system.

Table 4.2: Charges on the Different Subatomic Particles

Charge in Coulombs (C) Mass in Elementary Charges (e)

Electron -1.6021765 x 10~ 19 -1

Proton 1.6021765 x 10~ 19 1

Neutron

Notice that whether you use Coulombs or elementary charge units, when you ignore the positive and
negative signs, the charge on the proton and the charge on the electron have the same magnitude.

Previously, you learned that negative and positive charges of equal magnitude cancel each other out. This
means that the negative charge on an electron perfectly balances the positive charge on the proton. In
other words, a neutral atom must have exactly one electron for every proton. If a neutral atom has 1
proton, it must have 1 electron. If a neutral atom has 2 protons, it must have 2 electrons. If a neutral
atom has 10 protons, it must have 10 electrons. You get the idea. In order to be neutral, an atom must
have the same number of electrons and protons, but what kinds of numbers are we talking about? That's
what we'll look at in the next section.

• Most scientists don't refer to "elementary charges" as a unit. Nevertheless, if you treat elementary
charges just like you'd treat any another non-SI unit, like a pound (lb) or a foot (ft), they become a
lot easier to understand.

Atomic Number (Z) Identifies the Element

How do you tell the difference between a bike and a car? What about the difference between a car and a
unicycle? Take a look at Figure 4.13.

If you had to make a rule to distinguish between a unicycle, a bike, a car, what would it be? You can't use
color, because different cars can be different colors and, even worse, a car can be the same color as a bike
or unicycle. The same goes for weight. While most cars would weigh more than most bikes, which would
weigh more than most unicycles, that isn't always the case. In fact, that the little grey "Smart Car" in
Figure 4.13 probably weighs less than a large motorbike.

What you really need to distinguish between a car, a bike and a unicycle is a property that is the same
within each category, but different between the categories. A good choice would be the number of wheels.
All unicycles have one wheel, all bikes have two wheels, and all cars have four wheels. If you count wheels,
you will most likely never confuse a unicycle with a bike, or a bike with a car (even a motorbike with a
Smart Car!). In other words, if you know the number of wheels, you know which type of vehicle you're
dealing with.

Just as we can tell between cars, bikes, and unicycles by counting the number of wheels, scientists can tell
between different elements (remember, an element is a specific type of atom) by counting the number
of protons. If a vehicle has only one wheel, we know it's a unicycle. If an atom has only one proton, we
know it's a hydrogen atom or, said differently, it's an atom of the element hydrogen. Similarly, a vehicle
with two wheels is always a bike, just like an atom with two protons is always a helium atom, or an atom
of the element helium. When we count four wheels on a vehicle, we know it's a car, and when scientists
count four protons in an atom, they know it's a beryllium atom, or an atom of the element beryllium. The
list goes on: an atom with three protons is a lithium atom, an atom with five protons is a boron atom, an

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Unicycle

Bikes

Cars

Figure 4.13: A unicycle, three examples of cars, and 2 examples of bikes. Can you think of some rule that
might allow you to tell all unicycles, cars and bikes apart?

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atom with six protons is a carbon atom... in fact, we have names for atoms containing everything from 1
proton all the way up to 117 protons. So far, the maximum number of protons scientists have been able
to pack into a single atom is 117, and thus there are 117 known elements. (On Earth, only atoms with a
maximum of 92 protons occur naturally.)

Figure 4.14: You can't really distinguish between sulfur and gold based on color because both are yellowish.
You could say that gold was shiny, but then how would you tell the difference between gold and silicon?
Each element, however, does have a unique number of protons. Sulfur has protons, silicon has protons,
and gold has protons.

Since an atom of one element can be distinguished from an atom of another element by the number of
protons in its nucleus, scientists are always interested in this number, and how this number differs between
different elements (Figure 4.14). Therefore, scientists give this number a special name and a special
symbol. An element's atomic number (Z) is equal to the number of protons in the nuclei of any of
its atoms. The atomic number for hydrogen is Z = 1, because every hydrogen atom has 1 proton. The
atomic number for helium is Z = 2 because every helium atom has 2 protons. What's the atomic number
of carbon?

Mass Number (A) is the Sum of Protons and Neutrons

In the last section we learned that each type of atom or element has a specific number of protons. This
specific number was called the element's atomic number. Of course, since neutral atoms have to have one
electron for every proton, an element's atomic number also tells you how many electrons are in a neutral
atom of that element. For example, hydrogen has atomic number Z = 1. This means that an atom of
hydrogen has one proton, and, if it's neutral, one electron as well. Gold, on the other hand, has atomic
number Z = 79, which means that an atom of gold has 79 protons if it's neutral, and 79 electrons as well.
So we know the number of protons, and we know the number of electrons, but what about the third type
of subatomic particle? What about the number of neutrons in an atom?

The number of neutrons in an atom isn't important for determining atomic number; in fact, it doesn't even
tell you which type of atom (or which element) you have. The number of neutrons is important, though, if
you want to find a quantity known as the mass number (A). The mass number of any atom is defined
as the sum of the protons and neutrons in the atom:

mass numberA = (number of protons) + (number of neutrons)

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An atom's mass number is a very easy to calculate provided you know the number of protons and neutrons
in an atom

Example 1:

What is the mass number of an atom that contains 3 protons and 4 neutrons?
Solution:

(number of protons) = 3
(number of neutrons) = 4

mass number, A = (number of protons) + (number of neutrons)
mass number, A = (3) + (4)
mass number A = 7

Example 2:

What is the mass number of an atom of helium that contains 2 neutrons?
Solution:

(number of protons) = 2 Remember that an atom of helium always has 2 protons,
(number of neutrons) = 2

mass number, A = (number of protons) + (number of neutrons)
mass number, A = (2) + (2)
mass numberA = 4

Why do you think that the "mass number" includes protons and neutrons, but not electrons? You have
already learned that the mass of an electron is very, very small compared to the mass of either a proton
or a neutron (like the mass of a penny compared to the mass of a bowling ball). Counting the number of
protons and neutrons tells scientists about the total mass of an atom, but counting the number of electrons
would only confuse things.

Think of it this way - you're asked to lift a box containing some bowling balls and some pennies, but
the box has already been taped closed. Now, if you have to decide whether or not to get help lifting the
box, which would you prefer to know, the total number of bowling balls and pennies, or the just the total
number of bowling balls (Figure 4.15)? Suppose you were told only the number of bowling balls. If you
knew that there were 20 bowling balls in the box, you wouldn't lift the box on your own, but if you knew
that there was only 1, you probably would, even if that box contained 19 pennies that you didn't know
about. On the other hand, if, instead of being told the number of bowling balls, you were told the number
bowling balls and pennies, your decision would be more difficult. What if you were given the number 20?
That could mean 20 bowling balls and no pennies, or it could mean 1 bowling ball and 19 pennies. In
fact, it could even mean 20 pennies. Unfortunately, you would have no way of knowing what was meant
by the number 20. Certainly, you wouldn't choose to lift 20 bowling balls, but lifting 20 pennies would be
no problem. Just like you wouldn't care about the number of pennies in the box you were about to lift,
scientists don't care about the number of electrons when they calculate the mass number. That's why the
mass number is only the sum of the protons and neutrons in the atom.

Isotopes Have Varying Numbers of Neutrons

If you were reading the last section carefully, you'll already know that you can't use the number of neutrons
in an atom to decide which type of atom (or which element) you have. Unlike the number of protons, which

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Figure 4.15: Each of the boxes above contains a total of items. If you had to choose one to lift, though,
you'd want to know the number of bowling balls in the each box, not the total number of items in each
box. Obviously, you'd rather lift the box with bowling balls than the box with bowling balls.

is always the same in atoms of the same element, the number of neutrons can be different, even in atoms
of the same element. Atoms of the same element, containing the same number of protons, but different
numbers of neutrons are known as isotopes. Since the isotopes of any given element all contain the same
number of protons, they have the same atomic number (for example, the atomic number of helium is always
2). However, since the isotopes of a given element contain different numbers of neutrons, different isotopes
have different mass numbers. The following two examples should help to clarify this point.

Example 3:

What is the atomic number (Z), and the mass number of an isotope of lithium containing 3 neutrons. A
lithium atom contains 3 protons in its nucleus.

Solution:

atomic numberZ = (number of protons) = 3
(number of neutrons) = 3

mass number, A = (number of protons) + (number of neutrons)
mass number, A = (3) + (3)
mass number, A = 6

Example 4:

What is the atomic number (Z), and the mass number of an isotope of lithium containing 4 neutrons. A
lithium atom contains 3 protons in its nucleus.

Solution:

atomic numberZ = (number of protons) = 3
(number of neutrons) = 4

mass number, A = (number of protons) + (number of neutrons)
mass number, A = (3) + (4)
mass number, A = 7

Notice that because the lithium atom always has 3 protons, the atomic number for lithium is always Z = 3.
The mass number, however, is A = 6 in the isotope with 3 neutrons, and A = 7 in the isotope with 4
neutrons.

In nature, only certain isotopes exist. For instance, lithium exists as an isotope with 3 neutrons, and as
an isotope with 4 neutrons, but it doesn't exists as an isotope with 2 neutrons, or as an isotope with 5
neutrons. Scientists can make isotopes of lithium with 2 or 5 neutrons, but they aren't very stable (they
fall apart easily), so they don't exist outside of the laboratory.

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140

Atomic Mass is a Calculated Value

Of course, this whole discussion of isotopes brings us back to Dalton's Atomic Theory. According to
Dalton, atoms of a given element are identical. But if atoms of a given element can have different numbers
of neutrons, then they can have different masses as well! How did Dalton miss this? It turns out that
elements found in nature always exist as constant uniform mixtures of their naturally occurring isotopes.
In other words, a piece of lithium always contains both types of naturally occurring lithium (the type
with 3 neutrons and the type with 4 neutrons). Moreover, it always contains the two in the same relative
amounts (or "relative abundances"). In a chunk of lithium, 93% will always be lithium with 4 neutrons,
while the remaining 7% will always be lithium with 3 neutrons.

Unfortunately, Dalton always experimented with large chunks of an element - chunks that contained all of
the naturally occurring isotopes of that element. As a result, when he performed his measurements, he was
actually observing the averaged properties of all the different isotopes in the sample. Luckily, aside from
having different masses, most other properties of different isotopes are similar. As a result, the fact that
atoms of a given element aren't, strictly speaking, identical, isn't all that important for most chemistry
problems.

Knowing about the different isotopes is important, however, when it comes to calculating atomic mass. The
atomic mass of an element is the average mass of the masses of its isotopes and their relative percentages,
and is typically given in "atomic mass units" (u). (Remember that an "atomic mass unit" is a convenient
unit to use when studying atoms, because a proton is almost exactly 1.0 «). You can calculate the atomic
mass of an element provided you know the relative abundances the element's naturally occurring isotopes,
and the masses of those different isotopes. The examples below show how this is done.

Example 5:

Boron has two naturally occurring isotopes. In a sample of boron, 20% of the atoms are B-10, which is an
isotope of boron with 5 neutrons and a mass of 10 amu. The other 80% of the atoms are B-ll, which is
an isotope of boron with 6 neutrons and a mass of 11 amu. What is the atomic mass of boron?

Solution:

To do this problem, we will calculate 20% of the mass of B-10, which is how much the B-10 isotope
contributes to the "average boron atom." We will also calculate 80% of the mass of B-ll, which is how
much the B-ll isotope contributes to the "average boron atom."

Step One: Convert the percentages given in the question into their decimal forms by dividing each by 100:

20
Decimal form of 20% = = 0.20

100

80
Decimal form of 80% = = 0.80

Step Two: Multiple the mass of each isotope by its relative abundance (percentage) in decimal form:

20% of the mass of B - 10 = 0.20 x 10 amu = 2.0 amu
80% of the mass of B - 11 = 0.80 x 11 amu = 8.8 amu

Step Three: Find the total mass of the "average atom" by adding together the contributions from the
different isotopes:

Total mass of average atom = 2.0 u + 8.8 u = 10.8 u

The mass of an average boron atom, and thus boron's atomic mass, is 10.8 u.

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Example 6:

Neon has three naturally occurring isotopes. In a sample of neon, 90.92% of the atoms are Ne - 20, which
is an isotope of neon with 10 neutrons and a mass of 19.99 u. Another 0.3% of the atoms are Ne - 21,
which is an isotope of neon with 11 neutrons and a mass of 20.99 u. The final 8.85% of the atoms are
Ne - 22, which is an isotope of neon with 12 neutrons and a mass of 21.99 u. What is the atomic mass of
neon?

Solution:

To do this problem, we will calculate 90.9% of the mass of Ne-20, which is how much Ne-20 contributes
to the "average neon atom". We will also calculate 0.3% of the mass of Ne - 21 and 8.8% of the mass of
Ne - 22, which are how much the Ne - 21 isotope and the Ne - 22 isotope contribute to the "average neon
atom" respectively.

Step One: Convert the percentages given in the question into their decimal forms by dividing each by 100:

90 92
Decimal form of 90.92% = — = 0.9092

100

Decimal form of 0.30% = — = 0.0030

8 85
Decimal form of 8.85% = — = 0.0885

Step Two: Multiple the mass of each isotope by its relative abundance (percentage) in decimal form:

90.92% of the mass of Ne - 20 = 0.909 x 20.00 = 18.18 amu
0.3% of the mass of Ne - 21 = 0.003 x 21.00 = 0.063 amu
8.85% of the mass of Ne - 22 = 0.088 x 22.00 = 1.93 amu

Step Three: Find the total mass of the "average atom" by adding together the contributions from the
different isotopes:

Total mass of average atom = 18.18 amu + 0.06 amu + 1.93 amu = 20.17 amu
The mass of an average neon atom, and thus neon's atomic mass, is 20.17 amu.

Atomic Information in the Periodic Table

Most scientists don't want to have to calculate the atomic mass of an element every time they do an
experiment. Nor do they want to memorize the number of protons, or the atomic number, of each of
the 117 elements that have been discovered. As a result, this information is stored in the periodic table.
Figure 4.16 shows a periodic table that contains more detail than the periodic table you saw back in
Chapter 1.

Notice that each box still contains the symbol (a capital letter or a capital letter followed by a lower case
letter) for one of the elements, but now there are two new numbers that have been added to each square,
one number above the element's symbol, and another number below the element's symbol.

The number above the element's symbol in each square is the element's atomic number. Just as you learned
previously, hydrogen (symbol H) has atomic number Z = 1, helium (symbol He) has atomic number Z = 2,
lithium (symbol Li) has atomic number Z = 3, beryllium (symbol Be) has atomic number Z = 4, boron
(symbol B) has atomic number Z = 5, and carbon (symbol C) has atomic number Z = 6. The number
below the element's symbol in each square is the element's atomic mass. Notice that atomic mass of boron

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Periodic Table

1.01

4.00

6.94

9.01

10.1

12.0

14.0

16.0

19.0

20.2

23.0

24.3

Transition Metals

27.0

28.1

31.0

32.1

35.5

40.0

39.1

40.1

45.0

47. 9

£0. 9

52.0

54. 9

55. 9

58. 9

58.7

63. 6

65.4

69.7

72. 6

74. 9

79.0

79. 9

83.8

85.5

87.6

88.9

91.2

92.9

95.9

101

103

106

108

112

115

119

122

128

127

131

ATI

133

137

139

179

181

184

186

190

192

195

197

201

204

207

209

210

222

104

105

233

226

227

Lanthanides

Actinides

140

141

144

145

150

152

157

159

163

165

167

169

173

175

100

101

102

103

Am.

232

231

238

237

244

243

247

251

252

257

258

259

260

Figure 4.16: A periodic table showing both the atomic number (Z) of each element and the mass number
(A) of each element.

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(symbol B) is 10.8, which is what we calculated in example 5, and the atomic mass of neon (symbol Ne)
is 20.18, which is what we calculated in example 6. Observe how compactly the periodic table stores and
presents a large amount of information about each element. Take time to notice that not all periodic
tables have the atomic number above the element's symbol and the mass number below it. If you are ever
confused, remember that the atomic number (Z) should always be the smaller of the two, while the atomic
mass should always be the larger of the two. (The average mass must include both the number of protons
(Z) and the average number of neutrons).

You may listen to Tom Lehrer's humorous song "The Elements" with animation at The Element Song
(http : //www . privatehand . com/flash/elements . html)

Lesson Summary

Electrons are a type of subatomic particle with a negative charge. As a result, electrons repel each

other, but are attracted to protons.

Protons are a type of subatomic particle with a positive charge. As a result, protons repel each other,

but are attracted to electrons. Protons are bound together in an atom's nucleus as a result of the

strong nuclear force.

Neutrons are a type of subatomic particle with no charge (they're neutral). Like protons, neutrons

are bound into the atom's nucleus as a result of the strong nuclear force.

Protons and neutrons have approximately the same mass, but they are both much more massive than

electrons (approximately 2,000 times as massive as an electron).

The positive charge on a proton is equal in magnitude ("size when you ignore positive and negative

signs") to the negative charge on an electron. As a result, a neutral atom must have an equal number

of protons and electrons.

Each element has a unique number of protons. An element's atomic number (Z) is equal to the

number of protons in the nuclei of any of its atoms.

The mass number (A) of an atom is the sum of the protons and neutrons in the atom

mass number A = (number of protons) + (number of neutrons)

Isotopes are atoms of the same element (same number of protons) that have different numbers of

neutrons in their atomic nuclei.

An element's atomic mass is the average mass of one atom of that element. An element's atomic mass

can be calculated provided the relative abundances of the element's naturally occurring isotopes, and

the masses of those isotopes are known.

The periodic table is a convenient way to summarize information about the different elements. In

addition to the element's symbol, most periodic tables will also contain the element's atomic number

(Z), and element's atomic mass.

Review Questions

1. Decide whether each of the following statements is true or false.

(a) The nucleus of an atom contains all of the protons in the atom.

(b) The nucleus of an atom contains all of the neutrons in the atom.

(d) Neutral atoms of a given element must contain the same number of neutrons.

(e) Neutral atoms of a given element must contain the same number of electrons.

2. Match the subatomic property with its description.
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(a) electron - a. has an atomic charge of +le

(b) neutron - b. has a mass of 9.109383 x 10~ 28 grams

3. Arrange the electron, proton, and neutron in order of decreasing mass.

4. Decide whether each of the following statements is true or false.

(a) An element's atomic number is equal to the number of protons in the nuclei of any of its atoms.

(b) The symbol for an element's atomic number is (A).

(d) A neutral atom with A = 4 must have 4 electrons.

(e) An atom with 7 protons and 7 neutrons will have A = 14.

(f) An atom with 7 protons and 7 neutrons will have Z = 14.

(g) A neutral atom with 7 electrons and 7 neutrons will have A = 14.

5. Use the periodic table to find the symbol for the element with:

(a) 44 electrons in a neutral atom

(b) 30 protons

(d) an atomic mass of 14.007 amu

6. When will the mass number (A) of an atom be...

(a) bigger than the atomic number (Z) of the atom?

(b) smaller than the atomic number (Z) of the atom?

7. Column One contains data for 5 different elements. Column Two contains data for the same 5
elements, however different isotopes of those elements. Match the columns by connecting isotopes of
the same element.

Table 4.3:

Column One

Column Two

a. an atom with 2 protons and 1
neutron

b. a Be (beryllium) atom with 5
neutrons

c. an atom with Z = 6 and A =
13

d. an atom with 1 proton and
A = 1

e. an atom with Z = 7 and 7
neutrons

i. a C (carbon) atom with 6 neu-
trons

ii. an atom with 2 protons and 2
neutrons

iii. an atom with Z = 7 and A =
15

iv. an atom with A = 2 and 1
neutron

v. an atom with Z = 4 and 6
neutrons

Calculations:

8. Match the following isotopes with their respective mass numbers.

(a) an atom with Z = 17 and 18 neutrons - i. 35

(b) an H atom with no neutrons - ii. 4

(d) an atom with Z = 11 and 11 neutrons - iv. 23

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(e) an atom with 11 neutrons and 12 protons - v. 22
9. Match the following isotopes with their respective atomic numbers.

(a) a B (boron) atom with A = 10 - i. 8

(b) an atom with A = 10 and 6 neutrons - ii. 2

(d) an oxygen atom - iv. 4

(e) an atom with A = 4 and 2 neutrons - v. 5

10. Answer the following questions:

(a) What's the mass number of an atom that contains 13 protons and 13 neutrons?

(b) What's the mass number of an atom that contains 24 protons and 30 neutrons?

11. Answer the following questions:

(a) What's the mass number of the isotope of manganese (Mn) containing 28 neutrons?

(b) What's the mass number of the isotope of calcium (Ca) containing 20 neutrons?

12. Answer the following questions:

(a) What's the atomic number of an atom that has 30 neutrons, and a mass number of A = 70?

(b) What's the atomic number of an atom with 14 neutrons, if the mass number of the atom is
A = 28?

13. Answer the following questions:

(a) What's the mass number of a neutral atom that contains 7 protons and 7 neutrons?

(b) What's the mass number of a neutral atom that contains 7 electrons and 7 neutrons?

(d) What's the mass number of a neutral atom that contains 3 electrons and 4 neutrons

14. Answer the following questions:

(a) What element has 32 neutrons in an atom with mass number A = 58?

(b) What element has 10 neutrons in an atom with mass number A = 19?

15. Copper has two naturally occurring isotopes. 69.15% of copper atoms are Cu - 63 and have a mass
of 62.93 amu. The other 30.85% of copper atoms are Cu - 65 and have a mass of 64.93 amu. What
is the atomic mass of copper?

Vocabulary

electron A type of subatomic particle with a negative charge.

proton A type of subatomic particle with a positive charge. Protons are found in the nucleus of an atom.

neutron A type of subatomic particle with no charge. Neutrons are found in the nucleus of an atom.

the strong nuclear force The force that holds protons and neutrons together in the nucleus of the
atom. The strong nuclear force is strong enough to overcome the repulsion between protons.

atomic mass units (amu) A unit used to measure the masses of small quantities like protons, neutrons,
electrons and atoms. It is useful, because the mass of a proton is very close to 1.0 amu.

elementary charge (e) The magnitude of charge on one electron or one proton. You can treat elementary
charges as a unit of charge.

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atomic number (Z) An element's atomic number is equal to the number of protons in the nuclei of any
of its atoms.

mass number (A) The mass number of an atom is the sum of the protons and neutrons in the atom.

Image Sources

(1) Richard Parsons. . CC-BY-SA.

(2) Richard Parsons. . GNU-FLD.

(3) Richard Parsons. . CC-BY-SA.

(4) http : //en . wikipedia . org/wiki/Image : Christmas_Pudding . jpg. GNU-FDL.

(5) http://en.wikipedia.Org/wiki/Image:Gachalaemerald. jpg. Public Domain, CC-BY-SA.

(6) Richard Parsons. . CC-BY-SA.

(7) http : //en. wikipedia. or g / wiki/ Image: Bowl ingb all . jpg. Public Domain, Public Domain.

(8) .

(9) http://en.wikipedia.0rg/wiki/Image:Dalton_John_desk.jpg. Public Domain.

(10) Sharon Bewick. . CC-BY-SA.

(11) http: //en. wikipedia. org/wiki/Image :Native_gold_nuggets. jpg
http://en.wikipedia.0rg/wiki/Image:Silic0nCr0da. jpg. Public Domain, Public Domain, Public
Domain.

(12) http : //www . f lickr . com/photos/dottieday/536278579/. CC-BY-SA.

(13) http : //en . wikipedia . org/wiki/File : Hendrik_ter_Brugghen_-_Democritus .jpg. Public
Domain.

(14) http : //www . cheml . com/acad/webtext/pre/chemsci . html. CC-BY-SA.

(15) A portrait of J. J. Thomson.. Public Domain.

(16) http: //en. wikipedia. org/wiki/File :RacingBicycle-non. JPG
http : //en. wikipedia. or g/ wiki /Image: Roll s . arp . 850pix. jpg
http : //en. wikipedia. org/wiki/Image : HondaCBRlOOOF . jpg
http : //en. wikipedia. org/wiki/Image : Olpacecar . jpg

http://www.flickr.eom/photos/76074333@N00/269798915. GNU-FDL, GNU-FDL,Public
Domain,GNU-FDL,PublicDomain,CC-BY-SA.

147 www.cki2.0rg

Chapter 5

The Bohr Model

5.1 The Wave Form of Light

Lesson Objectives

The student will define the terms wavelength and frequency with respect to wave-form energy.
The student will state the relationship between wavelength and frequency with respect to electro-
magnetic radiation.

The student will state the respective relationship between wavelengths and frequencies of selected
colors on the electromagnetic spectrum.

Introduction

Our entire universe is made up of matter, which is anything that has mass and occupies space. You now
know that matter is composed of small building blocks known as atoms, and that these small building
blocks are composed of even smaller subatomic particles called protons, electrons and neutrons. Matter is
all around you and you can use the atomic, or subatomic, description of matter to understand anything
from the cells in your body to the planet Earth!

Any object that you can hold or touch is matter. But our universe contains something else - something
that you can't really touch, but that you can certainly see (in fact, you can't see without it!), and that you
can often feel. It isn't matter, because it doesn't have any mass, nor does it occupy any space. Still it's
fundamentally important to our everyday lives, and we most definitely have a name for it. Can you think
of what it is? If you haven't guessed by now, the answer is light.

Think about it for a minute - can you really talk about light using any of the ideas that we've considered so
far in our study of matter and the universe? Light doesn't have any mass. Light doesn't occupy any space
either. Try sticking your hand into the beam of light shining out of a flashlight. Your hand goes straight
through as if there was nothing there! And yet there must be something there... how else can you explain
the "brightness" that you see? If you have trouble understanding light and trying to define exactly what
light is, you're not alone. Scientists had trouble explaining light too. In fact, we've only really understood
light for about 100 years.

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Wave Form Energy

The wave model of energy can be partially demonstrated with waves in a rope. Suppose we tie one end of
a rope to a tree and hold the other end at a distance from the tree such that the rope is fully extended.

Figure 5.1: Wave in a rope.

If we then jerk the end of the rope up and down in a rhythmic way, the end of the rope we are holding
goes up and down. When the piece of rope we are holding goes up and down, it pulls on the neighboring
part of the rope which then also goes up and down. The up and down motion will be passed along to
each succeeding part of the rope so that after a short time, the entire rope will contain a wave as shown in
Figure 5.1. The red dotted line in the figure shows the undisturbed position of the rope before the wave
was initiated. The humps above the undisturbed line are called crests and the dips below the undisturbed
position are called troughs. It is important for you to recognize that the individual particles of the rope
DO NOT move horizontally. The particles of rope only move up and down and if the wave is allowed to
dissipate, all the particles of rope will be in exactly the same position they were in before the wave started.
Each hump in the rope moves horizontally from the person to the tree but the particles of rope only move
vertically. The energy that is put into the rope by jerking it up and down also moves horizontally from
the person to the tree. The feeling that parts of the rope are moving horizontally is a visual illusion.

If we jerk the rope up and down with a different rhythm, the wave in the rope will change its appearance in
terms of crest height, distance between crests, and so forth, but the general shape of the wave will remain
the same. We can characterize the wave in the rope with a few measurements. An instantaneous photo of
the rope will freeze it so we can indicate some of the characteristic values.

The distance from one crest to the next crest is called the wavelength of the wave (Figure 5.2). You
could also measure the wavelength from one trough to the next or, in fact, between any two identical
positions on successive waves. The symbol used for wavelength is the Greek letter lambda, A. The distance
from the maximum height of a crest to the undisturbed position is called the amplitude of the wave. We
could measure a velocity for the wave if we measure how far horizontally a crest travels in a unit of time.
The unit for velocity would be the normal meters/second. We also need to determine a very important
characteristic of waves called frequency. If we choose an exact position along the path of the wave and
count how many crests pass the position per unit time, we would get a value for frequency. In everyday
life, frequency values are often expressed as "cycles /second" or "waves/second" but when you try to use

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wavelength

amplitude

Figure 5.2: Characteristics of waves.

these units in calculations, the word "cycles" or "waves" will not cancel. The proper unit for frequency has
seconds in the denominator and "1" in the numerator. It is simple 1/s or s' 1 . This unit has been named
"Hertz." Frequencies are often expressed in Hertz but when you are plugging numbers into mathematical
formulas and wish to keep track of units, it is best to express frequency in units of s^ 1 . The symbol used
for frequency is the Greek letter nu, v. Unfortunately, this Greek letter looks a very great deal like an
italicized v. You must be very careful reading equations to be sure whether they are representing velocity,
v, or frequency, v. To avoid this problem, this material will use a lower case / as the symbol for frequency.
The velocity, wavelength, and frequency of a wave are related as indicated by the formula, v = fA. If the
wavelength is expressed in meters and the frequency is expressed in s , then multiplying the wavelength
times the frequency will yield m/s, which is the unit for velocity.

Electromagnetic Waves

Electromagnetic radiation (light) waves are somewhat like waves in a rope . . . except without the rope.
Light waves do not need a medium through which to travel. They can travel through a vacuum, which is
obvious since they come to us from the sun. The energy of an electromagnetic wave travels in a straight
line along the path of the wave just as did the energy in a rope wave. The moving light has associated
with it an oscillating electric field and an oscillating magnetic field. This means that along the straight
line path of the wave, there exists a positive electric field that will reach maximum positive charge, then
slowly collapse to zero charge, and then expand to a maximum negative charge. Along the path of the
electromagnetic wave, this changing electric field repeats its oscillating charge over and over again. There
is also a changing magnetic field that oscillates from maximum north pole field to maximum south pole
field. When scientists try to draw a picture to represent this concept, they use the same picture of a wave
that was used for rope waves and water waves.

wavelength

amplitude

You should not allow yourself to think that the light travels in this weaving pattern. The light travels along
the straight red line that represents the undisturbed position. For an electromagnetic wave, the crests and
troughs represent oscillating fields, not the path of the light. We can still characterize light waves by their

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150

wavelength, frequency, and velocity but these values will be significantly different numerically from water
and rope waves.

At some point along the path of the electromagnetic wave, the electric field will reach a maximum value
(crest) and then, as the electromagnetic wave continues to move along its straight line path, the electric
field will decrease, through zero, increase to a maximum trough, collapse back to zero again, and then
expand to another maximum crest. We can measure along the path of the wave, the distance the wave
travels between one crest and the succeeding crest and this distance will be the wavelength of the wave.
The frequency of electromagnetic waves are determined in the same way as the frequency of a rope wave,
that is, the number of full cycles that pass a point in a unit of time. The velocity of electromagnetic waves
(in a vacuum) is the same for all waves regardless of frequency or wavelength. Every electromagnetic wave
has a velocity of 3.00 x 10 8 m/s in a vacuum. The velocity of electromagnetic waves in air is slightly
less than in a vacuum but so close that we will use the value for the velocity. The speed of light in a
vacuum is symbolized by a lower case c. The relationship for the velocity, wavelength, and frequency of
electromagnetic waves is c = Af.

In rope waves and water waves, the amount of energy possessed by the wave is related to the amplitude
of the wave; more energy is put into the rope if the end of the rope is jerked higher and lower. But, in
electromagnetic radiation, the amount of energy possessed by the wave is related only to the frequency of
the wave. In fact, the frequency of an electromagnetic wave can be converted directly to energy in Joules
by multiplying by a conversion factor. The conversion factor is called Planck's constant and is equal to
6.63 x 10~ 34 J • s. Sometimes, the unit for Planck's constant is given as Joules/Hertz but you can work
that out to see the units are the same. The equation for the conversion of frequency to energy is E = hf,
where E is the energy in Joules, h is Planck's constant in J • s, and / is the frequency in s^ 1 .

The Electromagnetic Spectrum

Electromagnetic waves have an extremely wide range of wavelengths, frequencies, and energies. The highest
energy form of electromagnetic waves are gamma rays and the lowest energy form (that we have named)
are radio waves.

Roys

X-Rays

Ultraviolet
Rays

Shortwave

Infrared
Rays

Redan

1 x ID" 14 1 x ID -12
Wavelength (in meters!

x 10"* 1 x 10"

1 x 10* 1 x 10 4

4 x 10

5 x 10

6x ID" 7
Wavelength (in meters)

7 x 10

High Energy

Low Energy

Figure 5.3: The Electromagnetic Spectrum.

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On the far left of Figure 5.3 are the highest energy electromagnetic waves. These are called gamma
rays and can be quite dangerous, in large numbers, to living systems. The next lower energy form of
electromagnetic waves are called x-rays. Most of you are familiar with the penetration abilities of these
waves. They can also be dangerous to living systems. Humans are advised to limit as much as possible the
number of medical x-rays they have per year. Next lower, in energy, are ultraviolet rays. These rays are
part of sunlight and the upper end of the ultraviolet range can cause sunburn and perhaps skin cancer. The
tiny section next in the spectrum is the visible range of light . . . this section has been greatly expanded
in the bottom half of the figure so it can be discussed in more detail. The visible range of electromagnetic
radiation are the frequencies to which the human eye responds. Lower in the spectrum are infrared rays
and radio waves.

The light energies that are in the visible range are electromagnetic waves that cause the human eye to
respond when those frequencies enter the eye. The eye sends a signal to the brain and the individual
"sees" various colors. The highest energy waves in the visible region cause the brain to see violet and as
the energy decreases, the colors change to blue, green, yellow, orange, and red. When the energy of the
wave is above or below the visible range, the eye does not respond to them. When the eye receives several
different frequencies at the same time, the colors are blended by the brain. If all frequencies of light strike
the eye together, the brain sees white and if there are no visible frequencies striking the eye, the brain sees
black. The objects that you see around you are light absorbers - that is, the chemicals on the surface of
the object will absorb certain frequencies and not others. Your eyes detect the frequencies that strike your
eye. Therefore, if your friend is wearing a red shirt, it means the dye in that shirt absorbs every frequency
except red and the red frequencies are reflected. If your only light source was one exact frequency of blue
light and you shined it on a shirt that was red in sunlight, the shirt would appear black because no light
would be reflected. The light from fluorescent types of lights do not contain all the frequencies of sunlight
and so clothes inside a store may appear to be a slightly different color than when you get them home.

Lesson Summary

• One model of light is that of wave-form electromagnetic radiation.

• Light, in wave form, is characterized by its wavelength, A, frequency, /, and velocity, c.

• The unit for wavelength is meters and the unit for frequency is either s^ 1 or Hertz.

• The full spectrum of electromagnetic radiation has radio waves as its lowest energy, lowest frequency,
longest wavelength end and gamma rays as its highest energy, highest frequency, shortest wavelength
end.

• The colors we see for an object are the blending of all the frequencies of light reflected by the object.

Review Questions

1. Choose the correct word in for the following statement. Blue light has a (longer or shorter) wavelength
than red light.

2. Choose the correct word in for the following statement. Yellow light has a (higher or lower) frequency
than blue light.

3. Choose the correct word in for the following statement. Green light has a (larger or smaller) energy
than red light.

4. If "light A" has a longer wavelength than "light B", then "light A" has "light

B".

A. a lower frequency than
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B. a higher frequency than

C. the same frequency as

5. If "light C" has a shorter wavelength than "light D", then "light C" has "light

D".

A. a larger energy than

B. a smaller energy than

C. the same energy as

6. If "light E" has a higher frequency than "light F", then "light E" has -

"light F".

A. a longer wavelength than

B. a shorter wavelength than

C. the same wavelength as

7. If "light G" has a higher frequency than "light H", then "light G" has -

"light H".

A. a larger energy than

B. a smaller energy than

C. the same energy as

8. If "light J" has larger energy than "light K", then "light J" has "light

K".

A. a shorter wavelength than

B. a longer wavelength than

C. the same wavelength as

9. Which of the following statements is true?

A. The frequency of green light is higher than the frequency of blue light and the wavelength of green

light is longer than the wavelength of blue light.

B. The frequency of green light is higher than the frequency of blue light and the wavelength of green

light is shorter than the wavelength of blue light.

C. The frequency of green light is lower than the frequency of blue light and the wavelength of green light

is shorter than the wavelength of blue light.

D. The frequency of green light is lower than the frequency of blue light and the wavelength of green light

is longer than the wavelength of blue light.

E. The frequency of green light is the same as the frequency of blue light and the wavelength of green

light is shorter than the wavelength of blue light.

10. As the wavelength of electromagnetic radiation increases:

A. its energy increases.

B. its frequency increases.

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C. its speed increases.

D. more than one of the above statements is true.

E. none of the above statements is true.

11. List three examples of electromagnetic waves.

12. Why do white objects appear white?

13. Name the colors present in white light in order of increasing frequency.

14. Why do objects appear black?

Vocabulary

crest High point in a wave pattern (hill).

trough Low point in a wave pattern (valley).

amplitude of a wave The 'height' of a wave. In light waves, the amplitude is proportional to the
brightness of the wave.

frequency of a wave (v) The 'number' of waves passing a specific reference point per unit time. The
frequency of a light wave determines the color of the light.

hertz (Hz) The SI unit used to measure frequency. One Hertz is equivalent to 1 event (or one full wave
passing by) per second.

wavelength (A) The length of a single wave from peak to peak (or trough to trough). The wavelength of
a light wave determines the color of the light.

electromagnetic spectrum A list of all the possible types of light in order of decreasing frequency, or
increasing wavelength, or decreasing energy. The electromagnetic spectrum includes gamma rays,
X-rays, UV rays, visible light, IR radiation, microwaves and radio waves.

5.2 The Dual Nature of Light

Lesson Objectives

• Explain the double-slit experiment and the photoelectric effect.

• Explain why light is both a particle and a wave.

• Use and understand the formula relating a light's velocity, frequency, and wavelength, c = f A

• Use and understand the formula relating a light's frequency and energy, E = hf.

Introduction

Developing a theory to explain the nature of light was a difficult task. An acceptable theory in science
is required to explain ALL the observations made on a particular phenomenon. Light, appears to have
two different sets of behaviors under different circumstances. As you will see, sometimes light behaves
like wave-form energy and sometimes it behaves like an extremely tiny particle. It required the very best
scientific minds from all over the world to put together a theory to deal with the nature of light.

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The Difficulties of Denning Light Solved by Einstein and Planck

Why was light so hard to understand? Part of the problem was that when scientists performed different
experiments on light, they got conflicting results. Some experiments suggested that light was like a wave,
while others suggested that light was like a particle! Let's take a look at what's meant by "like a wave"
and "like a particle".

Figure 5.4: Water diffracting into circular waves as it passes through the small opening between two rocks.

Since you're probably familiar with water waves, we'll use water to explain wave behavior. Whenever a
water wave is forced through a small opening, such as the space between the two rocks in Figure 5.4, it
spreads out into a circular shape through a process known as diffraction. If several of these circular waves
run into each other, they can interfere with one another and produce interesting patterns in the water.

peaks

troughs

Figure 5.5: Patterns formed by colliding diffraction waves.

Figure 5.5 shows some of these patterns. Look carefully at the red line which defines a cross-section of
the pattern (portion (a)). That same cross-section is blown up in portion (b), where you can clearly see
how it is composed of alternating "peaks" and "troughs." The peaks are actually extra high points in the
waves (hills), while the troughs are extra low points (valleys).

Imagine how surprised scientists were when they shone light through two narrow slits in a solid plate and
saw a similar pattern of peaks (bright spots) and troughs (dark spots) on the wall opposite the plate.
Obviously, this proved that light had some very wave-like properties. In fact, by assuming that light was
a wave and that it diffracted through the two narrow slits in the plate, just like water waves diffract when
they pass between rocks, scientists were even able to predict where the bright spots would occur! Figure
5.6 shows how the results of the "double-slit" experiment could be understood in terms of light waves.

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I ^s bright =pots appear where
predicted, assuming light

travels as a wave

Figure 5.6: Waves of light also diffract when they pass through narrow slits.

What was also obvious from the double-slit experiment was that light could not be understood as a particle.
Imagine rolling a particle (like a marble) through a small opening. Would you expect it to diffract? Of
course not! You'd expect the marble to role in a straight line from the opening to the opposite wall,
as shown in Figure 5.7. Light traveling as a particle, then, should make a single bright spot directly
across from each slit opening. Since that wasn't what scientists observed, they knew that light couldn't be
composed of tiny particles.

I !

where bright spots would appear if
light traveled as a particle

where bright spots

j, actually appear

> where bright spots would appear if
light traveled as a particle

Figure 5.7: The bright spots formed when light shines through narrow slits are not straight through the
openings.

The wave theory of light seemed to work, at least for the double-slit experiment. Remember, though, that
according to the scientific method, a theory should be tested with further experiments to make sure that
it's accurate and complete. Unfortunately, the next experiment that scientists performed suggested that
light was not a wave, but was, instead, a stream of particles! By shining light on a flat strip of metal,
scientists found that they could knock electrons off of the metal surface. They called this phenomenon
the photoelectric effect, and they called electrons that were bumped off photoelectrons. Why did the
photoelectric effect prove that light wasn't a wave? The problem was that the number of photoelectrons
produced by a beam of light didn't depend on how bright the light was, but instead depended on the light's
color. To see why this was so important, we need to talk a little bit more about waves and light waves in
particular.

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Suppose you were sitting on a pier looking out at the Atlantic Ocean. Which do you think would be more
likely to knock you off the pier, a huge tidal wave or a wave like the type you might find on a calm day
at the beach? Obviously, the tidal wave would have a better chance of knocking you off the pier, and
that's because the tidal wave has more energy as a result of its bigger amplitude (amplitude is really just
another name for the 'height' of the wave). The energy of a wave depends on its amplitude, and only on
its amplitude. What does amplitude mean in terms of light waves? It turns out that in light waves, the
amplitude is related to the brightness of the light - the brighter the light, the bigger the amplitude of the
light wave. Now, based on what you know about tidal waves and piers, which do you think would be better
at producing photoelectrons, a bright light, or a dim light? Naturally, you'd think that the bright light
with its bigger amplitude light waves would have more energy and would therefore knock more electrons
off... but that's NOT the case.

It turns out that bright light and dim light knock exactly the same number of electrons off a strip of metal.
What matters, instead of the brightness of the light, is the color of the light (Figure 5.8). Red light
doesn't produce any photoelectrons, while blue light produces a lot of photoelectrons. Unlike brightness,
which depends on the amplitude of the light waves, color depends on their frequency (for a discussion of
frequency, see the box below).

AAAAAAAAAAAA,
AAAAAAAAAAAAAAAA/

Figure 5.8: The wavelengths of the different colors in the visible light spectrum.

This made the photoelectric effect very puzzling to scientists because they knew that the energy of a wave
doesn't depend on its frequency, only on its amplitude. So why did frequency matter when it came to
photoelectrons? It was all rather mysterious.

What is frequency? Frequency can be a difficult concept to understand, but it's really just a measure of
how many times an event occurs in a given amount of time. In the case of waves, it's the number of waves
that pass by a specific reference point per unit time. Figure 5.9 shows two different types of waves, one
red and one blue. Notice how, in a single second (one full turn of the clock hand), 4 red waves pass by the
dotted black line while 16 blue waves pass by the same reference point. We say that the blue waves have a
higher frequency than the red waves. The SI unit used to measure frequency is the Hertz (Hz). One Hertz
is equivalent to one event (or one full wave passing by) per second.

How does the frequency of the light affect the length of the light waves? Take a close look at Figure 5.9
again. What do you notice about the lengths of the blue and red waves? Obviously, the blue waves (higher
frequency) have a shorter wavelength, while the red waves (lower frequency) have a longer wavelength.
This has to be true, provided that the waves are traveling at the same speed. You can tell that the red
and blue waves are traveling at the same speed, because their leading edges (marked by a red dot and a
blue dot respectively) keep pace with each other. All light waves travel at the same speed.

The explanation of the photoelectric effect began with a man named Max Planck. Max Planck wasn't
actually studying the photoelectric effect himself. Instead, he was studying something known as black-
body radiation. Black-body radiation is the light produced by a black object when you heat it up (think,

157 www.ckl2.org

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Figure 5.9: Red and blue light have different wavelengths but travel at the same speed.

for example, of a stove element that glows red when you turn it on). Like the photoelectric effect, scientists
couldn't explain black-body radiation using the wave theory of light either. Max Planck, however, realized
that black-body radiation could be understood by treating light like a stream of tiny energy packets (or
particles). We now call these packets of energy "photons" or "quanta", and say that light is quantized.

Albert Einstein applied the theory of quantized light to the photoelectric effect and found that the energy
of the photons, or quanta of light, did depend on the light's frequency. In other words, all of a sudden
Einstein could explain why the frequency of a beam of light and the energy of a beam of light were related.
That made it a lot easier to understand why the number of photoelectrons produced by the light depended
on the light's color (frequency). The only assumption that Einstein needed to make was that light was
composed of particles.

Wait! Sure the particle theory of light explained black-body radiation and the photoelectric effect, but
what about the double-slit experiment? Didn't that require that light behave like a wave? Either the
double-slit experiment was wrong, or else the photoelectric effect and black-body radiation were wrong.
Surely, light had to be either a wave or a particle. Surely, it couldn't be both. Or could it? Albert Einstein
suggested that maybe light wasn't exactly a wave or a particle. Maybe light was both. Albert Einstein's
theory is known as the wave-particle duality of light, and is now fully accepted by modern scientists.

Light Travels as a Wave

You just learned that light can act like a particle or a wave, depending entirely on the situation. Did
you ever play with Transformers when you were younger? If you did, maybe you can answer the following
question: are Transformers vehicles (cars and aircraft) or are they robots? It's kind of a stupid question,
isn't it? Obviously Transformers are both vehicles and robots. It's the same with light - light is both a
wave and a particle.

Even though Transformers are both vehicles and robots, when they want to get from one place to another
quickly, they usually assume their vehicle form and use all of their vehicle properties (like wheels or airplane
wings). Therefore, if you were trying to explain how a Transformer sped off in search of an enemy, you'd
probably describe the Transformer in terms of its car or aircraft properties.

Just as it's easiest to talk about Transformers traveling as vehicles, it's easiest to talk about light traveling
as a wave. When light moves from one place to another, it uses its wave properties. That's why light
passing through a thin slit will diffract; in the process of traveling through the slit, the light behaves like
a wave. Keeping in mind that light travels as a wave, let's discuss some of the properties of its wave-like

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motion.

First, and most importantly, all light waves travel, in a vacuum, at a speed of 299,792,458 m/s (or
approximately 3.00 X 10 8 m/s). Imagine a tiny ant trying to surf by riding on top of a light wave (Figure
5.10). Provided the ant could balance on the wave, it would move through space at 3.00 x 10 8 m/s.

Figure 5.10: An ant surfing a light wave.

To put that number into perspective, when you go surfing at the beach, the waves you catch are moving
at about 9 m/s. Unlike light waves, though, which all travel at exactly the same speed, ocean waves travel
at different speeds depending on the depth of the ocean, the temperature, and even the wind!

Previously, you learned that light can have different frequencies and different wavelengths. You also learned
that because light always travels at the same speed 3.00 x 10 8 m/s, light waves with higher frequencies
must have smaller wavelengths, while light waves with lower frequencies must have longer wavelengths.
Scientists state this relationship mathematically using the formula

c / x A

speed of light(3.00 X 10 m/s) frequency(//z or s~ ) wavelength (m)

where c is the speed of light, 3.00 x 10 8 m/s, / is the frequency and A is the wavelength. Remember that
the unit we use to measure frequency is the Hertz (Hz), where 1 Hertz (Hz) is equal to 1 per second, s' 1 .
Wavelength, since it is a distance, should be measured in the SI unit of distance, which is the meter (m).
Let's see how the formula can be used to calculate the frequency or the wavelength of light.

Example 1:

What is the frequency of a purple colored light, if the purple light's wavelength is 4.45 X 10~ 7 m?

speed of light, c = 3.00 x 10 8 m/s You always know the speed of light, even if the question doesn't give it
to you.

wavelength, A = 4.45 x 10~ 7 m

c =fx A

(3.00 x 10 8 m/s) =fx 4.45xl0 _7 m

To solve for frequency, /, divide both sides of the equation by 4.45 x 10~ 7 m.

3.00 x 10 8 m/s _ 4.45 x 10~ 7 m
4.45 x 10- 7 m ~ X 4.45 x 10~ 7 m
6.74 xlO 14 j -1 = fx (1)

f = 6.74 x 10 14 Hz

The frequency of the purple colored light is 6.74 X 10 14 Hz.
Example 2:

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What is the frequency of a red colored light, if the red light's wavelength is 650 nm?
speed of light, c = 3.00 x 10 8 m/s

A = 650 nm

A = (650 nm)(l X 10~ 9 m/nm) = 6.50 X 10 -7 m

c =fx A

(3.00 x 10 8 m/s) =fx 6.50xl0 _7 m

To solve for frequency, /, divide both sides of the equation by 6.50 x 10~ 7 m.

3.00 x 10 8 m/s _ 6.50 x 10~ 7 m
6.50 x 10- 7 m ~ X 6.50 x lO" 7 m
4.61 x 10 14 s~ l = f x (1)

f=4.61xl0 14 Hz

The frequency of the red colored light is 4.61 x 10 14 Hz.

Notice that the wavelength in Example 1, 4.45 X 10~ 7 m, is smaller than the wavelength in Example
2,6.50 X 10 -7 m, while the frequency in Example 1, 6.74 X 10 14 Hz is bigger than the frequency in Example
2, 4.61 x 10 14 Hz. Just as you'd expect, a small wavelength corresponds to a big frequency, while a big
wavelength corresponds to a small frequency. (If you're still not comfortable with that idea, take another
look at Figure 6 and convince yourself of why this must be so, provided the waves travel at the same
speed.) Let's take a look at one final example, where you have to solve for the wavelength instead of the
frequency.

Example 3:

Scientists have measured the frequency of a particular light wave at 6.10 x 10 14 Hz. What is the wavelength
of the light wave?

speed of light, c = 3.00 x 10 8 m/s

frequency, / = 6.10 X 10 14 Hz

f = 6.10 x 10 s~ (To do dimensional analysis, it is easiest to change Hertz to per secondl Hz = ls~ )
c = f x A
3.00 x 10 8 m/s = 6.10 x 10 14 s' 1 x A

To solve for wavelength, A, divide both sides of the equation by 6.10 x 10 14 s^ 1 .

3.00 x 10 8 m/s 6.10 x lO 14 ^ 1

6.10 xlO 14 ^ 1 6.10 x 10 14 *- 1
4.92xl0~ 7 m = (l)xl

A = 4.92 x 10" 7 m

The wavelength of the light is 4.92 X 10 7 m (or 492 nm, if you do the conversion).
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Light Consists of Energy Packets Called Photons

We have already seen how the photoelectric effect proved that light wasn't completely wave-like, but
rather, had particle-like properties too. Let's return to our comparison between Transformers and light.
Transformers travel as vehicles; however, when Transformers battle each other, they fight as robots, not as
cars and planes. The situation with light is similar. Light may travel as a wave, but as soon as it strikes an
object and transfers its energy to that object, the light behaves as if it's made up of tiny energy packets,
or particles, called photons.

Remember, the energy of a wave depends only on the wave's amplitude, but not on the wave's frequency.
The energy of a photon, or a light "particle", however, does depend on frequency. The relationship between
a photon's energy and a photon's frequency is described mathematically by the formula

E = h X /

energy(/) Planck's constant(/z = 6.63 X 10~ / • s) frequency(//z or s~ )

where E is the energy of the photon, h is Planck's constant (which always has the value h = 6.63xl0 -34 J- s,
and / is the frequency of the light. The SI unit for nergy is the Joule (J); the SI unit for frequency is the
Hertz (or per second, s^ 1 ); the SI unit for Planck's constant is the Joule-second ( J • s). Although this
equation came from complex mathematical models of black-body radiation, its meaning should be clear -
the larger the frequency of the light beam, the more energy in each photon of light.

Example 4:

What is the energy of a photon in a stream of light with frequency 4.25 X 10 14 Hz?

Planck's constant, 6.63 x 1(T 34 J • s

frequency, 4.25 x 10 14 Hz

/ = 4.25 x 10 14 s -1 (To do dimensional analysis, it is easiest to change Hertz to per second, 1 Hz = 1 s" 1 )

E h x /

E = 6.63 x 10~ 34 / • s x 4.25 x lO^V 1

E = 2.82 x KT 19 /

The energy of a photon of light with frequency 4.25 x 10 14 s _1 is 2.82 x 10~ 19 J.

Example 5:

What is the frequency (in Hz) of a beam of light if each photon in the beam has energy 4.44 X 10 -22 J?

Planck's constant, 6.63 x 10~ 34 J • s

energy, E = 4.44 X 1CT 22 J

E h X /

4.44 x KT 22 ./ = 6.63 x 1(T 34 / • s x /

To solve for frequency (f), divide both sides of the equation by 6.63 x 1CT34 J • s

4.44 x KT 22 7 6.63 x 1(T 34 / • s
— x f

6.63 x 10- 34 7 • s 6.63 x lO" 34 / • s J

6.70 xlO 11 ^ 1 = (1) x /

/ = 6.70 x 10 ll Hz

The frequency of the light is 6.70 X 10 Hz

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The Electromagnetic Spectrum

When scientists speak of light in terms of its wave-like properties, they are often interested in the frequency
and wavelength of the light. One way that we, as humans, can distinguish between light beams of different
frequencies (and thus different wavelengths) is to use their colors. Light that is reddish colored has large
wavelengths and small frequencies, while light that is bluish colored has small wavelengths and large
frequencies. Humans can't, however, see all types of light. We can only see visible light. In fact, if the
light's wavelength gets too small, the light becomes invisible to our eyes. We call this light ultraviolet
(UV) radiation. Similarly, if the light's wavelength gets too large, the light also becomes invisible to our
eyes. We call this light infrared (IR) radiation.

Believe it or not, there are types of light with wavelengths even shorter than those of ultraviolet radiation.
We call these types of light X-rays and gamma rays. We can use X-rays to create pictures of our bones,
and gamma rays to kill bacteria in our food, but our eyes can't see either (you can see the picture that an
X-ray makes, but you can't actually see the X-ray itself). On the other side of the spectrum, light with
wavelengths even longer than those of infrared radiation are called microwaves and radio waves. We
can use microwaves to heat our food, and radio waves to broadcast music, but again, our eyes can't see
either.

wsmms
Roys

X-ftays

Ultraviolet
Rays

Infrared
Rays

Rader

Shortwave

1 x ID" 14 1 x ID" 12
Wavelength (in meters!

1CT 4 1 x 10" S

1 x 10 E 1 x 10 4

4 x 10" 7

High Enangy

5 x 10

6 x 10'
Wavelength (in meters)

7 x 10"

Low Energy

Figure 5.11: The electromagnetic spectrum.

Scientists summarize all the possible types of light in what's known as the electromagnetic spectrum.
Figure 5.11 shows a typical electromagnetic spectrum. As you can see, it's really just a list of all the
possible types of light in order of increasing wavelength. Notice how visible light is right in the middle of
the electromagnetic spectrum. Since light with a large wavelength has a small frequency and light with
a small wavelength has a large frequency, arranging light in order of "increasing wavelength," is the same
as arranging light in order of "decreasing frequency." This should be obvious from Figure 5.11 where, as
you can see, wavelength increases to the right (decreases to the left), while frequency increases to the left
(decreases to the right).

Unlike wavelength and frequency, which are typically shown on the electromagnetic spectrum, energy is
rarely included. You should, however, be able to predict how the energy of the light photons changes
along the electromagnetic spectrum. Light with large frequencies contains photons with large energies,

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162

while light with small frequencies contains photons with small energies. Therefore energy, like frequency,
increases to the left (decreases to the right).

Lesson Summary

• When waves pass through narrow openings, they spread out into a circular shape through a process
known as diffraction.

When circular waves interact, they produce predictable patterns of peaks and troughs.
When light is passed through two narrow slits, the light appears to interact in a manner similar to
two circular waves spreading out from the slits. This suggests that light diffracts into circular waves
when it passes through the slits and that these circular waves interact with each other. As a result,
many scientists believed that light was wave-like.

Shining light on a flat strip of metal knocks electrons off of the metal surface through what is known
as the photoelectric effect.

The number of photoelectrons produced by a beam of light depends on the color (wavelength) of the
light but not on the brightness (amplitude) of the light.

Since the energy of a wave should depend on the amplitude of the wave, scientists couldn't understand
why a brighter light didn't knock more photoelectrons off of the metal. This led them to question of
whether light was truly wave-like.

Together Max Planck and Albert Einstein explained the photoelectric effect by assuming that light
was actually a stream of little particles, or packets of energy known as photons or quanta.
Scientists now believe that light is both a wave and a particle - a property which they term the
wave-particle duality.

Light travels as a wave. The speed of a light wave is always c = 3.00 x 10 8 m/s. The frequency, /,
and wavelength, A, of a light wave are related by the formula c = f A.

Light gives up its energy as a particle or photon. The energy (E) of a photon of light is related to
the frequency, /, of the light according to the formula E = hf.

The relationship between the frequency, the wavelength, and the energy of light are summarized in
what's known as the electromagnetic spectrum. The electromagnetic spectrum is a list of light waves
in order of increasing wavelength, decreasing frequency, and decreasing energy.

Review Questions

1. Decide whether each of the following statements is true or false:

(a) Light always behaves like a wave.

(b) Light always behaves like a particle.

(d) Light travels like a wave and gives up its energy like a particle.

2. Which of the following experiments suggested that light was a wave, and which suggested that light
was a particle?

(a) the double-slit experiment

(b) the photoelectric effect

3. Fill in each of the following blanks.

(a) The brightness of a beam of light is determined by the of the light wave.

(b) The color of a beam of light is determined by the of the light wave

(frequency is also an acceptable answer)

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4. What is the name of the quantity depicted by each of the arrows in the diagram below (Figure
5.12)? (Source: Sharon Bewick. CC-BY-SA)

A AAA,

VU/v

Figure 5.12

5. Consider light with a frequency of 4.4 x 10 14 Hz. What is the wavelength of this light?

6. What is the frequency of light with a wavelength of 3.4 X 10 -9 m?

7. What is the frequency of light with a wavelength of 575 nm?

8. What is the energy of a photon in a beam of light with a frequency of 5.66 X 10 8 Hz?

Vocabulary

diffraction The tendency of a wave to spread out in a circular shape when passed through a small
opening.

double-slit experiment When light is passed through two narrowly separated openings (slits), the light
produces a resulting pattern of peaks and troughs that suggests that light behaves like a wave.

photoelectric effect The process whereby light shone on a metal surface knocks electrons (called pho-
toelectrons) off of the surface of the metal.

black-body radiation Light produced by a black object when the object is heated.

photon or quanta of light A tiny particle-like packet of energy.

wave- particle duality of light Einstein's theory, which concluded that light exhibits both particle and
wave properties.

5.3 Light and the Atomic Spectra

Lesson Objectives

• Distinguish between continuous and discontinuous spectra.

• Recognize that white light is actually a continuous spectrum of all possible wavelengths of light.

• Recognize that all elements have unique atomic spectra.

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Introduction

We now know how light can act as a wave or a particle, depending on the situation. You might wonder,
though, why a chemistry textbook would waste a whole lesson on light. Light, like matter, is part of the
universe, but chemists aren't responsible for studying the entire universe. Chemists are responsible for
studying chemicals. What does light have to do with chemicals? Why do chemists need to know about
light?

It turns out that scientists can actually learn a lot about chemicals by observing how they interact with
light. Different chemicals behave differently when struck with a beam of light. In fact, the same chemical
will interact differently with differently colored beams of light. To understand what light can tell us about
different chemicals, though, we must first look at the electromagnetic spectrum a little more carefully.

Continuous Spectra Compared to Discontinuous Spectra

If you have a class that requires a lot of work, you might find yourself saying something like, "I'm contin-
uously doing homework for this class". Think about what you mean by that. You probably mean that the
homework seems to be non-stop. Every time you finish one assignment, the teacher gives another, so you
never get a break. When scientists use the word continuous, it has a similar meaning. It means no
gaps, no holes, and no breaks of any kind.

Scientists don't use the word continuous to describe homework, but they do use it to describe electro-
magnetic spectra (spectra is just the plural word for spectrum). In the last section, you learned that an
electromagnetic spectrum was a list of light arranged in order of increasing wavelength. A continuous
electromagnetic spectrum, then, includes every possible wavelength of light between the wavelength at the
beginning of the list and the wavelength at the end. If you find that definition confusing, consider the
following example.

Suppose you have a continuous spectrum that begins with light at a wavelength of 500 nm, and ends
with light at a wavelength of 600 nm. Because it's continuous, that spectrum contains light with any
wavelength between 500 nm and 600 nm. It contains light with a wavelength of 550 nm. It contains light
with a wavelength of 545 nm. It contains light with a wavelength of 567.3 nm. It even contains light with
a wavelength of 599.99999 nm. Write down any number (including a number with decimal places) that
is bigger than 500 and smaller than 600. A continuous electromagnetic spectrum between 500 nm and
600 nm will include light with a wavelength equal to the number you've written down.

As shown in the last section, within the visible range of the electromagnetic spectrum, a light's wavelength
corresponds to its color. Therefore, another way of defining a continuous spectrum in the visible range is
to say that it is a spectrum which contains every possible color between the color at the beginning of the
list and the color at the end. Figure 5.13 shows several examples of continuous spectra in the visible light
range. The first continuous spectrum starts with a deep indigo blue and ends with red. Notice how the
colors in this spectrum change smoothly all the way from indigo to red. There are no gaps, or missing
colors. The same is true of the second continuous spectrum. The second spectrum again starts with a
deep indigo blue, but this time ends with yellow. Once more the colors in the spectrum change smoothly
without any gaps or holes, which makes the spectrum continuous. The third spectrum is also continuous,
only this time it starts with the color green and ends with the color orange.

Not all electromagnetic spectra are continuous. Sometimes they contain gaps or holes. Scientists call
electromagnetic spectra that contain gaps or holes discontinuous. Let's reexamine spectra that start at
500 nm and end at 600 nm. Discontinuous spectra in this range will include light with some, but not
all wavelengths of light greater than 500 nm and less than 600 nm. A discontinuous spectrum might, for
example, only contain light with wavelengths of 500 nm, 523 nm and 600 nm. Obviously you can think

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Figure 5.13: Several examples of continuous spectra in the visible range.

of many numbers that lie between 500 and 600 which aren't included in that list (534 is one example).
Therefore, the spectrum is discontinuous. A different discontinuous spectrum between 500 nm and 600 nm
might contain every wavelength of light except 533 nm. In this case, almost every wavelength of light is
included, but since 533 nm is missing, the spectrum is still discontinuous.

Figure 5.14: Several examples of discontinuous spectra in the visible range.

Figure 5.14 shows several examples of discontinuous spectra in the range of visible light. Again, since
the wavelength of a beam of light corresponds to its color, you can clearly see when an electromagnetic
spectrum in the visible range is discontinuous - there will be colors missing! In the first example, only
a few shades of green are missing from the middle of the spectrum. Nevertheless, the missing shades of
green make the spectrum discontinuous. The next two examples in Figure 5.14 have even bigger gaps of
missing color, so it's even more obvious that they are discontinuous.

The concept of a continuous spectrum compared to a discontinuous spectrum may seem a little silly. So
what if one spectrum contains every possible wavelength, while another skips wavelengths here and there!

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166

Why does that matter? To understand the importance of continuous and discontinuous spectra, we have
to look a little more closely at the ways in which light interacts with matter, and how those interactions
can actually produce electromagnetic spectra.

Each Element Has Its Own Spectrum

Light from the sun is a continuous spectrum. In other words, when you go to the beach and sunbathe,
you are bombarded by light beams of every different wavelength in the electromagnetic spectrum. Certain
wavelengths bounce off your skin, while others interact in ways that lead to a tan or even sun burn. For
example, you've probably seen sunscreens that offer UV protection. Ultraviolet light has wavelengths
that are smaller than those of visible blue light. In addition to the white light that we see, the continuous
spectrum of light from the sun contains UV light, and light with that range of wavelengths can be dangerous
to human skin cells.

A sunbather trying to avoid tan lines and possibly too much UV radiation. (Source: http : //en . wikipedia .
org/wiki/ Image : Sunbathing, jpg. Public Domain)

Of course, when you're lying in the sun on the beach, you don't actually see a rainbow shining down on
you, do you? Instead, you see white light. As a result, you might be skeptical and find it hard to believe
that sunlight forms a continuous spectrum. Surely, the beams of light coming from the sun don't contain
every possible wavelength of light... surely, they only contain the wavelengths of light corresponding to the
'color' white. That argument might seem logical to you, but you've fallen into a common trap - white
isn't a color. If you take a careful look at the electromagnetic spectrum, what you'll notice is that there is
no 'white' light in the visible range. It turns out that white light does not come from light of any specific
wavelength or range of wavelengths. Rather, in order for our eyes to see white, they must actually receive
light of every wavelength in the entire visible spectrum.

When sunlight passes through water, the white sunlight is spread out so that you can actually see the
entire spectrum of brightly colored light composing it. This is what we know as a rainbow. (Source:
http : //en . wikipedia . org/wiki/File : Surf ing_Rainbow . JPG . GNU-FDL)

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White is not a color, and there is no light with a wavelength corresponding to 'white'. Instead, white light
is formed when light of every wavelength in the visible spectrum is mixed together.

Remember that by passing white light through a prism, you were able to split the light into a rainbow,
revealing all of the different colors, or different wavelengths of light that make up white light.

Now that we know where to find continuous spectra (light from the sun or any other source of pure white
light), let's discuss where and when discontinuous spectra appear in our world. By passing an electric
current or an electric spark through certain types of matter, it's possible to make that matter glow. Neon
lights are one common example of this phenomenon. When an electric current travels through neon gas,
the neon glows bright orange.

When an electric current is passed through neon gas, the gas glows orange,
wikipedia . org/wiki/ Image : NeTube . jpg . CC-BY-SA)

(Source: http://en.

When an electric current is passed through argon gas, the gas glows blue. (Source: http : //en . wikipedia .
org/wiki /Image: ArTube.jpg. CC-BY-SA)

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168

Neon isn't the only gas that lights up when electricity passes through it. Electricity causes argon to glow
blue and helium to glow pink. In fact, electricity causes every element in the entire periodic table to glow
with a distinct color. As you might have guessed, the light from a glowing sample of neon or argon is
very different from the light shining down from the sun. Unlike sunlight, which is white, elements such
as neon and argon glow in colors, which means that the light they emit (or send forth) is missing certain
wavelengths - if it wasn't, it would appear white to our eyes.

Remember how sunlight spread out into a rainbow, or a continuous spectrum, when you passed it through
a prism? Well, when you pass light from a sample of glowing hydrogen through a prism, it doesn't spread
out into a continuous spectrum. Instead, it spreads out into a discontinuous spectrum, with only four
lines of colored light. A similar thing happens when you pass the light from a sample of glowing neon, or
argon, or even sodium through a prism. Instead of getting a continuous spectrum, you get a discontinuous
spectrum composed of a series of colored lines. The particular series of colored lines that you get out of
any specific element is called the element's atomic spectrum or emission spectrum. Each element
has an emission spectrum that is characteristic to that element. In other words, the emission spectrum
from sodium is always the same and is different than the emission spectrum from any other element, like
calcium or helium, or gold.

The emission spectrum for hydrogen. (Source: http://en.wikipedia.org/wiki/Atomic_emission_
spectrum. CC-BY-SA)

The emission spectrum for sodium. (Source: http : //en . wikipedia . org/wiki/Atomic_emission_spectrum .
CC-BY-SA)

Emission spectra are important to scientists for two reasons. First, because an element's emission spectrum
is characteristic of the element, scientists can often use emission spectra to determine which elements are
present or absent in an unknown sample. If the emission spectrum from the sample contains lines of
light that correspond to sodium's emission spectrum, then the sample contains sodium. You may have
heard or read about scientists discussing what elements are present in some distant star, and after hearing
that, wondered how scientists could know what elements are present in a place no one has ever been.
Scientists determine what elements are present in distant stars by analyzing the light that comes from
stars and finding the atomic spectrum of elements in that light. If the emission spectrum from the sample
contains lines of light that correspond to helium's emission spectrum, then the sample contains helium.
Second, and perhaps more importantly, the existence of atomic spectra and the fact that atomic spectra
are discontinuous, can tell us a lot about how the atoms of each element are constructed. In general, an
element's atomic spectrum results from the interaction between the electrons and protons within an atom
of that element. The relationship between atomic spectra and the components of the atom will be the
topic of the next lesson.

Lesson Summary

• When scientists use the word continuous, they mean something with no holes, no gaps, and no breaks.

• A continuous electromagnetic spectrum contains every wavelength between the wavelength on which
the spectrum starts and the wavelength on which the spectrum ends.

• A discontinuous electromagnetic spectrum is a spectrum that contains gaps, holes, or breaks in terms

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of the wavelengths that it contains.

• Light from the sun and, in fact, any pure white light source, produces light that contains a continuous
spectrum of wavelengths.

• White is not a color of light itself, but rather, results when light of every other color is mixed together.

• When an electric current or an electric spark is passed through an element, the element will give off
a colored glow. This glow is actually composed of light from a discontinuous spectrum that is unique
to the each and every element.

• We call the discontinuous spectrum produced by passing an electric current through an element the
element's atomic spectrum or emission spectrum.

• Atomic spectra can be used to identify elements. They also tell us a lot about the nature of matter.

Review Questions

1. Decide whether each of the following spectra is continuous or discontinuous. (Source: Sharon Bewick.
CC-BY-SA)

2. Decide whether each of the following statements is true or false.

(a) White light has a wavelength of 760 nm.

(b) Sodium's atomic spectrum is an example of a discontinuous spectrum.

3. Fill in each of the following blanks with the words 'must', 'may' or 'does not.'

(a) A continuous spectrum between 300 nm and 565 nm contain light with a wavelength

of 356 nm.

(b) A continuous spectrum between 1000 cm and 1.500 m contain light with a wavelength

of 1.234 m.

of 300 mm.

4. Choose the correct word in each of the following statements.

(a) A continuous spectrum between 532 nm and 894 nm contains light of every wavelength (greater
than/less than) 532 nm and (greater than/ less than) 894 nm

(b) A discontinuous spectrum between 532 nm and 894 nm contains light of every wavelength
between 532 nm and 894 nm (including/except for) light with a wavelength of 650 nm

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5. What is another name for an atomic spectrum?

6. When an electric current is passed through neon, it glows .

7. When an electric spark is passed through argon, it glows .

8. When an electric current is passed through helium, it glows .

9. Suppose you had a solution which contained both dissolved hydrogen and dissolved sodium. If an
electric current was passed through this solution, how many lines would you see in the emission
spectrum?

10. LEDs, or light emitting diodes, produce light by passing an electric current through a mixture of
different atoms (or molecules) and then using their combined emission spectra to light up a room, or
a string of Christmas tree lights. Why are white LEDs difficult and expensive to make?

Vocabulary

continuous electromagnetic spectrum A spectrum that contains every possible wavelength of light
between the wavelength at the beginning of the list and the wavelength at the end. In the visible
range of light, it is a spectrum which contains every possible color between the color at the beginning
of the list and the color at the end.

discontinuous electromagnetic spectrum A spectrum that includes some, but not all of the wave-
lengths in the specified range. In the visible spectrum there are gaps or missing colors.

pure white light A continuous spectrum of all possible wavelengths of light

atomic spectrum (emission spectrum) A unique, discontinuous spectrum emitted by an element
when an electric current is passed through a sample of that element

5.4 The Bohr Model

Lesson Objectives

• Define an energy level in terms of the Bohr model.

• Find the energy of a given Bohr orbit using the equation E n = = ^ £ .

• Discuss how the Bohr model can be used to explain atomic spectra.

Introduction

In the last lesson, you learned that atoms of different elements produce different atomic spectra when they
are struck by an electric spark or an electric current. This phenomenon is really rather puzzling. Why do
atoms emit light when they are exposed to an electric current? Why is the emitted light only at specific
wavelengths? Why do different elements have different atomic spectra? Surely these atomic spectra must
tell us something about the atoms that they came from, but what does it all mean? These were the types of
questions that scientists were asking in the early 1900s when a Danish physicist named Niels Bohr (Figure
5.15) became interested in atomic spectra and the nature of the atom.

Bohr Used Atomic Spectra to Develop His Model

We have learned that light travels as a wave and gives up its energy as a particle. As a result, the
wavelength (or color) of a light is related to the light's frequency which is, in turn, related to the light's

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Figure 5.15: A photograph of Niels Bohr.

energy. All of these relationships were summarized in the electromagnetic spectrum. Remember the
smaller the wavelength of the light (blue end of the visible spectrum), the larger the frequency and the
larger energy. Similarly, the larger the wavelength of the light (red end of the visible spectrum), the smaller
the frequency and the smaller the energy.

When Niels Bohr began thinking about the atom and atomic spectra in general, he was aware of the
wave-particle duality of light that had just been discovered. He knew that a specific wavelength (or color)
of light was related to the light's energy. As a result, Bohr realized that the lines of color appearing in
an element's atomic spectrum corresponded not only to those wavelengths of light but also, to the specific
frequencies and, more importantly, to the specific energies of light. The important question, then, was why
the atoms were only emitting light at very specific energies. Niels Bohr realized that this unusual result
could be explained by proposing what he termed energy levels. What are energy levels?

Electron Energy Is Quantized

You have learned that a rock held above the ground would release potential energy as it fell. Niels Bohr
realized that in order for atoms to release energy, there must be a similar 'falling' process going on inside the
atom. Since Bohr, like Rutherford, knew that the protons in the atom were bound up in the tiny nucleus,
the obvious sub-atomic objects that could be 'falling' inside the atom were the electrons. Therefore, Bohr
proposed a model in which electrons circled around the nucleus and, on occasion, fell closer to the nucleus,
releasing energy in the process. According to Bohr, the energy that came out of the atoms as their electrons
fell towards the nucleus appeared as light. This light, he argued, produced the atomic spectra that could
be seen whenever electric current was passed through an element.

Of course, none of Bohr's arguments thus far explain why only certain energies appear in each element's
atomic spectrum. If you hold a rock above the ground and drop it, it will release a specific amount of
energy. If you raise that rock higher by a tiny amount, it will release slightly more energy. If you lower
that rock by a tiny amount, it will release slightly less energy. In fact, it would seem you can get that rock
to release any quantity of energy that you'd like, just by raising and lowering it to different levels above

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ground. But with electrons, the situation is clearly different. With electrons, it seems as if there are only a
few levels from which electrons can fall. At least, that's what Bohr decided, and that's why he proposed
the existence of the atomic energy level.

According to Bohr, the electrons in an atom were only allowed to exist at certain energy levels. The
electrons could jump from a lower level to a higher level when they gained energy (which they could get
from a passing electric current, electric spark, heat, or light), and they could drop from a higher level to a
lower level when they lost energy (which they released in the form of light). Most importantly, though, the
electrons could not exist in between the allowed energy levels. Many people have compared Bohr's energy
levels to a set of stairs. Think about a child jumping up and down a set of stairs. The child can rest at
any one of the stairs and stay there. If he puts energy in, he can jump up to a higher stair. If he allows
himself to fall, he can drop to a lower stair. The child cannot, however, hover at a level in between two of
the stairs, just as an electron cannot hover in between two of the atom's energy levels.

Just as children cannot hover between two steps on a staircase, Bohr suggested that electrons cannot hover
between to energy levels in the atom. (Source: http://www.flickr.com/photos/veganstraightedge/
327066889/. CC-BY-SA)

Bohr developed his energy level model further using principles from physics. In physics, the energy of a
positive charge and a negative charge depends on the distance between them. Therefore, Bohr decided
that his energy levels must correspond to orbits, or circular paths centered around the nucleus of the
atom. Since an electron that was trapped in one of these orbits remained a constant distance from the
nucleus, Bohr stated that within an orbit an electron had a constant energy. Of course, the fact that only
certain energy levels were allowed, also meant that only certain orbits were allowed. Figure 5.16 shows a
schematic illustration of Bohr's model. In the diagram, each circle from n = 1 to n = 4 is an allowed orbit.

Notice how the negative electron, in any given orbit, is always the same distance from the positive nucleus
at any place on the orbit. As a result electrons within an orbit always have the same energy. When an
electron is hit by electricity, it can gain energy and can be bumped up to a higher energy orbit further
away from the nucleus. On the other hand, when an electron loses energy, it falls back down to a lower
energy orbit closer to the nucleus. The electron can never exist at distances in between allowed orbits.
Notice how this limits the number of different transitions that the electron can make. Because the electron
can only exist in certain orbits, and thus can have only certain energies, we say that the energy of the
electron is quantized.

Bohr Used a Formula to Determine Allowed Energy Levels

Although Bohr's descriptive model of atomic orbits gives a nice explanation as to why atoms should
have discontinuous atomic spectra, it's hard to test experimentally. Luckily, using advanced physics, the
Bohr model can be used to derive a mathematical expression for the energies of the allowed orbits in the
hydrogen atom. As you'll see in the next section, the energies of these orbits actually determine which

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Figure 5.16: A schematic illustration of the Bohr model of the atom.

wavelengths of light appear in hydrogen's atomic spectrum, meaning that Bohr's model can, in fact, be
tested experimentally. The equation predicting the energies of the allowed hydrogen orbits is:

Rydberg constant, R = 1.097 x 10 7 mr 1

Planck's constant, h = 6.63 x 1CT 34 J • s

speed of light, c = 3.00 x 10 8 m/s

-Rxhxc

Here E„ is the energy of the n orbit (in other words the energy of the 1 st , 2 , 3 , 4 , etc. orbit), R is the
Rydberg constant for hydrogen (which always has the value R = 1.097 X 10 7 m" 1 , h is Planck's constant
(which always has the value h = 6.63 x 10~ 34 J • s, c is the speed of light (which always has the value
c = 3.00 X 10 8 m/s, and n is the number of the orbit you are interested m.n can have any integer value
(no decimals!) from 1 to infinity, n = 1, 2, 3, 4 . . . infinity. As was shown in Figure 5.16, n = 1 is the orbit
closest to the atomic nucleus, n = 2 is the next orbit out, n = 3 is the one after that, and so on. In other
words, n increases with increasing distance from the nucleus. Let's take a look at several examples using
this equation.

Example 1:

What is the energy of the 3 allowed orbit in the hydrogen atom?

Planck's constant, 6.63 x 10~ 34 J • s

Rydberg constant for hydrogen R = 1.097 X 10 7 m~ 1 (Notice how you do not need to be given h, R or c,
since they are all constants)

speed of light, 3.00 x 10 8 m/s

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n = 3

174

-R x h x c
E —

(1.097 x 10 7 m _1 ) x (6.63 x 10 _34 J.s) x (3.00 x 10 8 m/s)

(3)2
(7.27 x 10- 27 J.s/m) x (3.00 x 10 8 m/s)

9
(2.18 xl0~ 18 J)
9

E„ = -2.42 x 10~ 19 J

(Be careful not to drop the negative sign! These energies should always be negative)

Example 2:

What is the energy of the 2nd allowed orbit in the hydrogen atom?

Planck's constant, h = 6.63 x 10~ 34 J • s

Rydberg constant for hydrogen, R = 1.097 X 10 m (Notice, again, how you do not need to be given h, R
or c, since they are all constants)

speed of light, c = 3.00 x 10 8 m/s

n = 2

-fixhxc

(1.097 x 10 7 m _1 ) x (6.63 x 10 _34 J.s) x (3.00 x 10 8 m/s)

(7.27 x 10- 27 J.s/m) x (3.00 x 10 8 m/s)
E n - -

(2.18 xlO -18 J)
E n - -

E n = -5.45 x 10" 19 J

(Again, be careful not to drop the negative sign!)

At the moment, these probably seem like fairly boring calculations, but in the next section, we'll see
how energy level calculations like the two above can actually be used to predict the atomic spectrum of
hydrogen. When scientists first did this in the early 1900s, they were amazed at how well this simple
equation predicted the colors of light emitted by hydrogen atoms.

Atomic Spectra Produced by Electrons Changing Energy Levels

Suppose you have $10 in your wallet when you go to the store, and you only have $4 when you come home.
How much money have you spent at the store? The answer is simple - you've obviously spent $6. With
electrons, the situation is similar. If they start with 10 units of energy, and fall down to 4 units of energy,
they lose 6 units of energy in the process. To state this mathematically, we write:

&Ef-*f — Ef — E(

Change in energy from initial state, i, to final state, f Energy in final state,/ Energy in the state, i

1T5 www.ckl2.org

When electrons 'lose' energy, though, they lose it by giving it off in the form of light. We have calculated
the energy of an electron in the 3 hydrogen orbit, -2.42 X 1CT 19 J, and the energy of an electron in the
2 hydrogen orbit, -5.45 X 10~ 19 J. An electron falling from the 3 hydrogen orbit to the 2 hydrogen
orbit begins with -2.42 x 10 -19 J, and ends with -5.45 x 10~ 19 J, thus it loses -3.03 x 10 -19 J in the process.
This can be worked out mathematically. (Don't worry too much about the fact that this lost energy has a
negative sign. Any time that an atom gains energy, the sign will be positive, and any time that an atom
loses energy, the sign will be negative).

initial state, i — 3

final state, / = 2

E f = E 2 = -5.45 x 1(T 19 J
E t = E 3 = -2.42 x 10~ 19 J

A£ , 3-»2 = £2 - £3

AEf^f = (-5.45 x 1(T 19 J) - (-2.42 x 1(T 19 J)

Remember, subtracting a negative number is the same as adding a positive number!

AEj^ f = (-5.45 x 1(T 19 J) + (2.42 x 1(T 19 J)
AEj^f = -3.03 x 10~ 19 J

All of the energy lost when an electron falls from a higher energy orbit to a lower energy orbit is turned into
light, thus when an electron falls from the 3 hydrogen orbit to the 2 hydrogen orbit, it emits a beam of
light with an energy of 3.03 x 10~ 19 J. You should now be able convert this energy into a wavelength (first
convert the energy into a frequency and then, convert that frequency into a wavelength ). The wavelength
turns out to be 656 nm and, amazingly, if you look at hydrogen's atomic spectrum, shown in Figure ??,
you'll notice that hydrogen clearly has a line of red light at exactly 656 nm! What's more, the green 486 nm
line in hydrogen's atomic spectrum corresponds to an electron falling from the 4 th hydrogen orbit to the
2 hydrogen orbit, the blue 434 nm line corresponds to an electron falling from the 5 th hydrogen orbit to
the 2 nd hydrogen orbit, and the purple 410 nm line corresponds to an electron falling from the 6 th hydrogen
orbit to the 2 nd hydrogen orbit. In other words, Bohr's model can be used to predict the exact wavelengths
of the four visible lines in hydrogen's atomic spectrum.

410 nm

434 nm

486 nm

656 nm

A short discussion of atomic spectra and some animation showing the spectra of elements you chose and an
animation of electrons changing orbits with the absorbtion and emission of light can be viewed at Spectral
Lines (http : //www . Colorado . edu/physics/2000/quantumzone/index . html) .

What about all of the other transitions that could occur between the allowed hydrogen orbits? For example,
where is the line that corresponds to an electron falling from the 4th orbit to the 3rd orbit? It turns out that
the energies of all the other transitions don't lie within range of energies included in the visible spectrum.
As a result, even though these transitions occur, your eyes can't see the light that's given off as a result.
Clearly, then, the predictions from Bohr's model give a perfect fit when compared to the hydrogen atomic
spectrum. There's only one small problem...

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176

Bohr's Model Only Worked for Hydrogen

If you take a look at the periodic table, what you'll notice is that hydrogen is a very special element.
Hydrogen is the first element, and thus, it only has one electron. It turns out that Bohr's model of the
atom worked very well provided it was used to describe atoms with only one electron. The moment that
the Bohr model was applied to an element with more than one electron (which, unfortunately, includes
every element except hydrogen), the Bohr model failed miserably.

Bohr's model failed because it treated electrons according to the laws of classical physics. Unfortunately,
those laws only apply to fairly large objects. Back when Bohr was developing his model, scientists were only
beginning to realize that the laws of classical physics didn't apply to matter as tiny as the electron. Electrons
are actually quantum objects, meaning that they can only be described using the laws of quantum physics.
Many of the differences between classical physics and quantum physics become particularly important when
two or more quantum objects interact. As a result, while Bohr's model worked for hydrogen, it became
worse and worse at predicting the atomic spectra for atoms with more and more electrons. Even helium,
with two electrons, was something of a disaster!

Bohr's model explained the emission spectrum of hydrogen which previously had no explanation. The
invention of precise energy levels for the electrons in an electron cloud and the ability of the electrons to
gain and lose energy by moving from one energy level to another offered an explanation for how atoms were
able to emit exact frequencies of light. Bohr calculated energies for the energy levels of hydrogen atoms
that yielded the exact frequencies found in the hydrogen spectrum. Furthermore, those same energy levels
predicted that hydrogen atoms would also emit frequencies of light in the infrared and ultraviolet regions
that no one had observed previously. The subsequent discovery that those exact frequencies of infrared
and ultraviolet light were present in the hydrogen spectrum provided even greater support for the ideas in
the Bohr model.

One of the problems with Bohr's theory was that it was already known that when electrons were accelerated,
they emitted radio waves. When you study physics, you will learn that acceleration applies to speeding
up, slowing down, and traveling in a curved path. When charged particles are accelerated, they emit
radio waves. In fact, that is how we create radio signals ... by forcing electrons to accelerated up and
down in an antenna. Scientists were creating radio signals in this way since 1895. Since Bohr's electrons
were supposedly traveling around the nucleus in a circular path, they MUST emit radio waves, hence lose
energy and collapse into the nucleus. Since the electrons in the electron cloud of an atom did not emit
radio waves, lose energy, and collapse into the nucleus, there was some immediate doubt that the electrons
could be traveling in a curved path around the nucleus. Bohr attempted to deal with this problem by
suggesting that the electron cloud contained a certain number of energy levels, that each energy level could
hold only a single electron, and that in ground state, all electrons were in the lowest available energy level.
Under these conditions, no electron could lose energy because there was no lower energy level available.
Electrons could gain energy and go to a higher energy level and then fall back down to the now open
energy level thus emitting energy, but once in ground state, no lower positions were open. This explained
why electrons circling the nucleus did not emit energy and spiral into the nucleus. Bohr did not, however,
offer an explanation for why only the exact energy levels he calculated were present, that is, what is there
about electrons in electron clouds that produce only a specific set of energy levels.

Another problem with Bohr's model was the predicted positions of the electrons in the electron cloud. If
Bohr's model was correct, the hydrogen atom electron in ground state would always be the same distance
from the nucleus. If we could take a series of photographic snapshots of a hydrogen electron cloud that
would freeze the position of the electron so we could see exactly where it was located at different times,
we still wouldn't know the path the electron followed to get from place to place, but we could see a few
positions for the electron. Such an image of electron positions would show the electron could actually be
various distances from the nucleus rather that at a constant distance.

ITT www.ckl2.org

A. ••>*•. B * • .

Figure 5.17: The electrons are NOT a constant distance from the nucleus.

If the electron circled the nucleus as suggested by Bohr's model, the electron positions would always be
the same distance from the nucleus as shown in Figure 5.17, portion A. In reality, however, the electron
is found at many different distances from the nucleus as in Figure 5.17, portion B. To solve all these
discrepancies, scientists would need a completely new way of looking at not just energy but at matter as
well.

The development of Bohr's model of the atom is a good example of the scientific method. It shows how
the observations of atomic spectra lead to the invention of a hypothesis about the nature of matter to
explain the observations. The hypothesis also made predictions about spectral lines that should exist in
the infrared and ultraviolet regions and when these observations were found to be correct, it provided even
more supportive evidence for the theory. Of course, further observations can also provide contradictory
evidence that will cause the downfall of the theory which also occurred with Bohr's model. Bohr's model
was not, however, a failure. It provided the insights that allowed the next step in the development of our
concept of the atom.

Bohr's Model Unacceptable

You have just seen that the Bohr model applied classical physics to electrons when electrons can only be
described using quantum mechanics. In addition, it turns out that Bohr's description of electrons as tiny
little objects circling the nucleus along fixed orbits is incorrect as well. Bohr was picturing an atom that
looked very much like a small solar system. The nucleus at the center of the atom was like the sun at the
center of the solar system, while the electrons circling the nucleus were like the planets circling the sun.
However, in quantum mechanics, electrons are thought of more like clouds rather than planets. Rather
than 'circling' the nucleus confined to orbits, electrons can seem to be everywhere at once, like a fog.

The fact that Bohr's model worked as well as it did for hydrogen is actually quite remarkable! Of course,
Bohr's fictional 'solar system atom' wasn't a random guess. Bohr had actually thought quite a bit about
what might be going on inside the atom, and his work marked the first major step towards understanding
where electrons are found in the atom. Therefore, despite the fact that the Bohr model wasn't entirely
correct, Niels Bohr was awarded a Nobel Prize for his theory in 1922. It turns out that a complete
description of the atom and atomic spectra, requires an understanding of quantum physics. Quantum
physics describes a bizarre world that behaves according to rules which only apply to very, very small
objects like electrons. Back when Bohr developed his model of the atom, scientists had never heard of
quantum physics. In fact, most scientists at the beginning of the 20th century thought that everything
could be described using classical physics. Even today, scientists still don't fully understand quantum
physics and the world that quantum physics describes.

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Lesson Summary

• Niels Bohr suggested that electrons in an atom were restricted to specific orbits around the atom's
nucleus.

• Bohr argued that an electron in a given orbit has a constant energy, thus he named these orbits
energy levels.

• When an electron gains energy (from an electric current or an electric spark), it can use this energy
to jump from a lower energy orbit (closer to the nucleus) to a higher energy orbit (farther from the
nucleus).

• When an electron falls from a higher energy orbit (farther from the nucleus) to a lower energy orbit
(closer to the nucleus) it releases energy in the form of light.

• White is not a color of light itself, but rather, results when light of every other color is mixed together

• In Bohr's model, electrons can only exist in certain orbits and thus, can only have certain energies.
As a result, we say that the energies of the electrons are quantized.

• Bohr used the formula E„ = = ^j^ to predict he energy level of an electron in the nth energy level (or
orbit) of a hydrogen atom.

• Because, the electron is only allowed to exist at certain energy levels according to the Bohr model,
there are only a few possible energies of light which can be released when electrons fall from one
energy level to another. As a result, the Bohr model explains why atomic spectra are discontinuous.

• The Bohr model successfully predicts the four colored lines in hydrogen's atomic spectrum, but it
fails miserably when applied to any atom with more than one electron. This is due to the differences
between the laws of classical physics and the laws of quantum physics.

• The Bohr model is no longer accepted as a valid model of the atom.

Review Questions

1. Decide whether each of the following statements is true or false:

(a) Niels Bohr suggested that the electrons in an atom were restricted to specific orbits and thus
could only have certain energies.

(b) Bohr's model of the atom can be used to accurately predict the emission spectrum of hydrogen.

(d) According to the Bohr model, electrons have more or less energy depending on how far around
an orbit they have traveled.

2. According to the Bohr model, electrons in an atom can only have certain, allowable energies. As a
result, we say that the energies of these electrons are .

3. The Bohr model accurately predicts the emission spectra of atoms with...

(a) less than 1 electron.

(b) less than 2 electrons.

(d) less than 4 electrons.

4. Consider an He + atom. Like the hydrogen atom, the He + atom only contains 1 electron, and thus
can be described by the Bohr model. Fill in the blanks in the following statements.

(a) An electron falling from the n = 2 orbit of He + to the n = 1 orbit of He + releases -

energy than an electron falling from the n = 3 orbit of He + to the n = 1 orbit of He + .

(b) An electron falling from the n = 2 orbit of He + to the n = 1 orbit of He + produces light with

a wavelength than the light produced by an electron falling from the n = 3 orbit of

He + to the n = 1 orbit of He + .

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frequency than the light produced by an electron falling from the n = 3 orbit of He +

to the n = 1 orbit of He + .

5. According to the Bohr model, higher energy orbits are located (closer to/further from) the atomic
nucleus. This makes sense since negative electrons are (attracted to/repelled from) the positive
protons in the nucleus, meaning it must take energy to move the electrons (away from/towards) the
nucleus of the atom.

6. According to the Bohr model, what is the energy of an electron in the first Bohr orbit of hydrogen?

7. According to the Bohr model, what is the energy of an electron in the tenth Bohr orbit of hydrogen?

8. According to the Bohr model, what is the energy of an electron in the seventh Bohr orbit of hydrogen?

9. If an electron in a hydrogen atom has an energy of -6.06 X 10~ 20 J, which Bohr orbit is it in?

10. If an electron in a hydrogen atom has an energy of -2.69 X 10~ 20 J, which Bohr orbit is it in?

11. If an electron falls from the 5 th Bohr orbital of hydrogen to the 3 rd Bohr orbital of hydrogen, how
much energy is released (you can give the energy as a positive number)?

12. If an electron falls from the Q th Bohr orbital of hydrogen to the 3 Bohr orbital of hydrogen, what
wavelength of light is emitted? Is this in the visible light range?

Vocabulary

Bohr energy level Distinct energies corresponding to the orbits (or circular paths) of electrons around
the atomic nucleus, according to Bohr's model of the atom

Bohr model of the atom Bohr's explanation of why elements produced discontinuous atomic spectra
when struck by an electric current. According to this model, electrons were restricted to specific
orbits around the nucleus of the atom in a solar system like manner.

classical physics The laws of physics that describe the interactions of large objects

quantum mechanics The laws of physics that describe the interactions of very small (atomic or sub-
atomic) objects. Also known as "wave mechanics" and "quantum physics".

Image Sources

(2
(3
(4
(5

(6
(7

(8
(9

Richard Parsons. . CC-BY-SA.

Sharon Bewick. A schematic illustration of the Bohr model of the atom. CC-BY-SA.
Sharon Bewick. An ant surfing a light wave.. CC-BY-SA.
A photograph of Niels Bohr.. Public Domain.
Richard Parsons. The Electromagnetic Spectrum.. CC-BY-SA.
http : //www . f lickr . com/photos/f ramesof mind/554402976/. CC-BY-SA.
Sharon Bewick. . CC-BY-SA.
Sharon Bewick. . CC-BY-SA.
http : //en . wikipedia . org/wiki/Image : Spectrum4websiteEval . svg. CC-BY-SA.

www.ckl2.org 180

(10) http : //en . wikipedia . org/wiki/Image : Gluehlampe_01_KMJ . jpg. GNU-FDL.

(11) Patterns formed by colliding diffraction waves.. CC-BY-SA.

(12) Sharon Bewick. . CC-BY-SA.

(13) http : //en . wikipedia. org/wiki/Image : Gluehlampe_01_KMJ .jpg. GNU-FDL.

(14) CK-12 Foundation. Wave in a rope.. CC-BY-SA.

(15) Richard Parsons. . CC-BY-SA.

(16) CK-12 Foundation. Characteristics of waves.. CC-BY-SA.

(17) http : //en . wikipedia . org/wiki/Image : Spectrum4websiteEval . svg. CC-BY-SA.

181 www.ckl2.org

Chapter 6

Quantum Mechanics Model of
the Atom

6.1 The Wave Particle Duality

Lesson Objectives

Explain the wave-particle duality of matter.

Define the de Broglie relationship and give a general description of how it was derived.

Use the de Broglie relationship to calculate the wavelength of an object given the object's mass and

velocity.

Introduction

In The Bohr Model of the Atom chapter you learned about the double-slit experiment. The double-slit
experiment proved that light passing through two closely spaced slits diffracted into circular waves, which
then interfered with each other and made interesting patterns on the opposite wall. From the double-slit
experiment, it was obvious that light behaved like a wave. But then along came a new set of results from
black body radiation and the photoelectric effect. Both of these experiments could only be understood
by assuming that light was a particle! Eventually, scientists had no choice but to accept the fact that
light was actually both a wave and a particle - hence the wave-particle duality of light. Now, you probably
didn't find the arguments in the previous chapter all that hard to accept. You probably thought to yourself
"Fine. If Einstein says light is a wave and a particle, I'll believe him. He's the super-genius scientist, and
I've never understood light anyhow."

Now let's talk about something that you probably do understand (or at least think you do). Let's talk
about matter. Before reading any further, answer the following question - "Does matter behave like a
wave, or a particle?" Obviously it behaves like a particle, right? Didn't we already decide that matter was
made up of tiny particles called atoms, and that those atoms were, themselves, made up of even tinier
particles called electrons, protons and neutrons? Don't worry. Everything that you've learned so far is
absolutely true. Atoms, electrons, protons and neutrons do behave like particles. But that's not the whole
story. Atoms, electrons, protons and neutrons also behave like waves! In other words, matter is just like
light in that it has both wave-like and particle-like properties.

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Electrons Were First Only Considered to Have Particle Proper-
ties

Do you remember how electrons were first discovered? It was all thanks to J. J. Thomson, with his clever
cathode ray tube experiment. Briefly, J. J. Thomson cut a small hole into the anode of a cathode ray tube.
This allowed cathode rays to pass through the anode and strike phosphor-coated glass on the other side.
Since the phosphor glowed when it was struck by the cathode rays J. J. Thomson could actually visualize
how the cathode ray traveled through the cathode ray tube.

What J. J. Thomson noticed, of course, was that the cathode rays hit the glass of the cathode ray tube
directly opposite the hole in the anode. In other words, the cathode rays traveled in a straight line from
the cathode, through the anode, to the glass at the end of the cathode ray tube. That sounds pretty much
like a particle, doesn't it? After all, if the electrons in the cathode ray were wave-like, wouldn't they have
diffracted when they passed through the hole in the anode? In that case, a huge glowing circle should have
appeared on the phosphor painted glass, rather than just a tiny glowing spot.

Early experiments didn't show any evidence of electron diffraction, so most scientists believed that electrons
were particles. It turned out, though, that electrons really do have wave-like properties. The wavelengths
were just too small to diffract in the cathode ray experiment.

Without diffraction, the electrons
make the phosphor glow directly
across from the hole in the anode.

calhod

phoptwr coatk>g

cathode

if the electrons had diffracted, they
would have coated a Sa*ge

glowing circle on (he phosphor.

In addition, by experimenting with magnets, J. J. Thomson had proven that the electron had mass. Up
until the 1920s, scientists had never observed wave-like behavior from an object with mass. In fact, you
probably can't think of anything with mass that exhibits obvious wave properties. Baseballs don't diffract;
bullets fired from neighboring rifles at a shooting range never form interference patterns.

You might ask about water waves, like the type you see in the ocean. Technically speaking, the water in

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the ocean doesn't behave like a wave. Instead, it's the energy traveling through the water that behaves
like a wave. Water may have mass, but the energy traveling through the water is 'mass-less.' So far we've
established the fact that you never see an object with mass exhibit wave-like properties. Why, then,
would anyone suggest that matter, even matter as small as an electron, could be described in terms of a
wave-particle duality?

What we know as ocean waves are actually energy waves passing through the water. (Source: http:
//en.wikipedia. org/wiki/Image: Waves_in_pacif ica_l . jpg. GNU-FDL)

de Broglie Proposed That Electrons are Particles and Waves

We never see matter behaving like a wave, but in 1924 a French graduate student named Louis de Broglie
(Figure 6.1) suggested that matter did in fact, have wave-like properties. It was a strange proposal on de
Broglie's part, but even stranger was the fact that no one laughed at him! Instead, de Broglie was awarded
a Nobel Prize in 1929. So how did de Broglie convince the scientists of his day to believe his theory? In
order to argue against de Broglie's wave-particle duality of matter, scientists would have had to argue with
Einstein - and no one argues with Einstein!

Figure 6.1: Louis de Broglie

Einstein is most famous for saying "mass is related to energy ". Of course, this is usually written out
www.ckl2.org 184

as an equation, rather than as words:
E = m x c x c

E = m x c 2

Speed of Light

(c = 3.00 s 10 s m/s)

Energy in
Joules

Mass in kg

(In The Science of Chemistry chapter, you were told that you didn't need to know where the equation
E — mc 2 comes from. You still don't need to know where it comes from, but you do need to know what it
means!) So Einstein was responsible for defining the amount of energy in any object with mass. Remember,
though, that Einstein was also responsible for proposing the wave-particle duality of light. In section 5.1.3
you learned that, because of the wave-particle duality of light, the energy of a wave can be related to the
wave's frequency by the equation:

E
1

= 1

Energy

Planck's Constant

(6.6

3 1 10 J4 Js>

Louis de Broglie looked at Einstein's first equation relating mass and energy. Then he looked at Planck's
equation, relating energy and frequency (remember, in The Bohr Model of the Atom chapter you learned
that frequency is a property associated with all waves, and is related to wavelength according to the
equation c = f A). After thinking for a while, de Broglie said to himself, "If mass is related to energy,
and energy is related to frequency, then mass must be related to frequency!" That's really a very
logical argument. It's like saying "If I'm related to my brother, and my brother is related to my
sister, then I must be related to my sister." Even though de Broglie's argument was logical, it led to a
very illogical result. If mass is related to frequency, then an object with mass must also have frequency.
In other words, an object with mass must have wave-like properties! This led de Broglie to propose the
wave-particle duality of matter. But where were these so called matter waves? Why hadn't anybody seen
them?

de Broglie Derived an Equation for the Wavelength of a Particle

Using Einstein's two equations, E = mc 2 and E = hf, along with the equation relating a wave's frequency
and its wavelength, c = A f , de Broglie was able to derive the following relationship between the wavelength
of an object to the object's mass:

li ^ Planck's Constant

X, = (6_63xl0" 3+ J.s)

I m x c

I I Ni

"W avelength (m)

X Speed of Light

Mass (kg) (c = 3.00 x 10 s m ; s)

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If you're good at math, you may be able to derive the exact same formula yourself. When de Broglie looked
at his equation, though, he realized that something was wrong. Why should the wavelength of an object
like an electron, or a baseball, depend on the speed of light? Neither baseballs nor electrons travel at the
speed of light - only light travels at the speed of light! The problem, of course, was that de Broglie had
derived his relationship using equations that applied to light, and not to matter. Luckily, the problem was
easy to fix. All de Broglie needed to do was replace the speed of light, c, with some general speed, v.

A, =

I m x v

Wavelength (m)

Planck's Constant
(6.63 xlO" 34 J-s)

Velocity (m/s)

Mass (kg)

That way, the de Broglie relationship between mass and wavelength could be applied to any thing, traveling
at any speed. It could be applied to light, traveling at the speed c = 3.00 x 10 8 m/s, it could be applied to
a baseball, traveling at the speed v = 45 m/s, and it could be applied to a marathon runner traveling at
the speed of v = 3 m/s. Let's try to find the wavelengths of a few common objects (Figure 6.2).

Figure 6.2: The wavelength for a typical baseball is far, far too small to be detectable, even in a laboratory!

Example 1:

One of the fastest baseballs ever pitched traveled at a speed of 46.0 m/s. If the average baseball has a
mass of 0.145 kg, what is the baseball's wavelength?

Planck's constant, h = 6.63 x 10~ 34 J.s

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186

(Notice that you always know Planck's constant, even if the question doesn't give it to you)
speed, v = 46.0 m/s
mass, m = 0.145 kg

h (6.63 x 10~ 34 J • s) 6.63 x 10~ 34 J • s ,„ J • s

9.94 x 10"

m X v (0.145 kg)(46.0 m/s) 6.67 kg • m/s kg • m/s

Dividing by a fraction is the same as multiplying by its reciprocal. This applies to units as well as numbers,
so dividing by ( kg/m • s) is the same as multiplying by ( s/kg • m).

^ = 9.94xl0- 35 ^^
kg • m

In order to account for units, you have to substitute basic units for compound units. The definition of a
Joule is, 17 = 1 kg • m 2 /s 2 . Substituting these basic units into the equation in place of Joules yields:

w kg • m 2 s • s

A = 9.94 x 10~ 35 -2— — x

kg • m

After cancelling units, the resultant unit is meters, the correct unit for wavelength.

A = 9.94 x 10~ 35 m

The wavelength for a typical stock car is far, far too small to be detectable. (Source: http://en.
wikipedia.org/wiki/Image:NASCAR_practice. jpg. Public Domain)

Example 2:

Stock cars typically race at around a speed of 77.0 m/s. If the average stock car has a mass of 1312 kg,
what is the stock car's wavelength?

Planck's constant, h = 6.63 x 10 -34 J.s

Again, you always know Planck's constant, even if the question doesn't give it to you)

speed, v = 77.0 m/s

mass, m = 1312 kg

h (6.63 x 10~ 34 J • s) 6.63 x 10~ 34 J • s „ r _ w in _ 39 J • s

A = = -^ — '— = — = 6.57 x 10"

mxv (1310 kg)(77.0 m/s) 100870 kg • m/s ' kg • m/s

18T www.cki2.0rg

-39 kg ' m x S-S \*n.™ T i„„ .•„ i„„„j ....^u i.„ ™2/„2

s 2 kg • m

/I = 6.57 x 10 ^ — x When Joules is replaced with kg • m /s

After cancelling units, the resultant unit is meters, the correct unit for wavelength.

A = 6.57 x 1(T 39 m

What do you notice about the wavelength of a typical baseball and the wavelength of a typical stock car?
They're both extremely small, aren't they? In fact, even the strongest microscope in the world today can't
see down to sizes like 9.94 x 10~ 35 m or 6.57 x 10 -39 m. Well, that explains why we've never seen a matter
wave - matter waves are just too small.

If matter waves are too small to see, then how did scientists prove that they exist? After all, the scientific
method requires experimental evidence before a theory is accepted - and certainly before a scientist is
awarded a Nobel Prize for that theory! Luckily, you can see matter waves. The only trick is to pick matter
with waves that are big enough to see. Take another look at the two example questions that we just did.
Notice how the only two factors influencing the wavelength of an object are the object's mass and the
object's speed. (Of course, you also need to know the value of Planck's constant, h = 6.63 X 1CT 34 J • s,
but since it's a constant, it doesn't change). In our two examples, the car was more massive than the
baseball, and it was traveling at a faster speed. What do you notice about the car's wavelength compared
to the baseball's wavelength? The car's wavelength was a lot smaller, wasn't it?

It turns out that the larger the mass of an object, and the faster the speed at which the object is traveling,
the smaller the object's wavelength. Similarly, the smaller the mass of an object, and the slower the speed
at which the object is traveling, the larger the object's wavelength. This is a direct consequence of de
Broglie's equation for the wavelength - because both mass, m, and speed, v, appear in the denominator
(the lower part of the fraction), they must both be big if you want a small wavelength, and small if you
want a big wavelength.

Obviously, in order to see matter waves experimentally, it would be best to have a big wavelength. That
means we need an object with a small mass and a slow speed. What's the smallest object that you know?
It's an electron of course! Let's see if an electron wave is large enough to detect in a laboratory

Example 3:

What is the wavelength of an electron traveling at 1.25xl0 5 m/s if the mass of the electron is 9.11xl0 -31 kg?

Planck's constant, h = 6.63 x 1CT 34 J • s

Remember, you always know Planck's constant, even if the question doesn't give it to you

speed, v = 1.25 x 10 5 m/s

mass, m = 9.11 x 1CT 31 kg

h (6.63 x 10~ 34 J • s) 6.63 x 1CT 34 J • s q J ■ s

1 - - - 5.82 x 10" 9

m xv (9.11 X 1CT 31 kg)(1.25 X 10 5 m/s) 1.14 X 1CT 25 kg • m/s ' kg • m/s

n kg • m 2 s • s iii

A = 5.82 x 10 = — x When Joules is replaced with kg • m/s .

s 2 kg • m

After cancelling units, the resultant unit is meters, the correct unit for wavelength.

A = 5.82 x 10~ 9 m

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Now that wavelength is bigger! All right, it's still not huge, but it's big enough that, in the 1920s, scientists
were able to find evidence of electron waves diffracting as electrons were forced through a thin metal film.
But let's return to one of our initial questions - if electrons have wave-like properties, why didn't J. J.
Thomson see electron diffraction in his cathode ray tube? The answer is simple - in that experiment, the
hole that the electrons had to pass through was just too big to cause diffraction. Waves only diffract when
their wavelengths are about the same size as the opening that they are forced through. Since electron
waves are extremely small, there's no way that they will diffract unless they are forced through extremely
small openings. As for your body when you walk through a door - well, if you're worried about your body
diffracting, use your mass and the speed at which you walk through the door to figure out your wavelength.
That should tell you about how narrow the door has to be to cause your body to diffract. From now on,
you can avoid doors of that size! (If you do the calculation properly, you should get a number around
1 x 10~ 35 and, of course, you couldn't fit through a door of that size).

Lesson Summary

• At first electrons were thought to behave only as particles; de Broglie stated that all matter has
wave properties and used Einstein's E = mc 2 formula and the formula E = hf to derive the formula:
A = h/(mv) to describe the wavelength A of an object with mass m traveling at speed v.

• Matter waves are impossible to detect for ordinary objects, like baseballs and cars, because they
are extremely small. The larger the mass, and the faster the speed of an object, the smaller its
wavelength.

• Scientists found evidence of electrons diffracting when forced through a thin metal film.

Review Questions

1. In the last chapter you learned that light has wave-like properties and particle- like properties. Can
you think of anything else that might have both wave-like properties and particle-like properties?

2. Decide whether each of the following statements is true or false.

(a) Einstein was the first scientist to propose matter waves.

(b) You can see baseballs diffract when you throw them.

(d) Most matter waves are very small, and that is why scientists didn't realize matter had wave-like
properties until the 1920s.

3. Choose the correct word in each of the following statements.

(a) The (more/less) massive an object is, the longer its wavelength is

(b) The (faster/slower) an object is traveling, the shorter its wavelength is

(c) A particle with a mass of 1.0 g has a (longer/ shorter) wavelength than a particle with a mass
of 3.0 g if both are traveling at the same speed

(d) A baseball moving at 10 m/s has a (longer/ shorter) wavelength than a baseball moving at 4 m/s

4. Choose the correct word in each of the following statements

(a) An electron has a (longer/shorter) wavelength than a proton if both are traveling at the same
speed.

(b) An electron wave has a (higher/lower) frequency than a proton wave if both particles are trav-
eling at the same speed.

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5. Choose the correct statement from the options below. The factors that influence an object's wave-
length are...

(a) Only the speed of the object

(b) Only the speed of light

(d) Only the mass of the object

(e) The speed of the object and the mass of the object

6. Choose the correct statement from the options below.

(a) Light behaves only like a wave, and matter behaves only like a particle

(b) Light behaves only like a wave, and matter behaves only like a wave

(d) Light behaves like a wave and like a particle, but matter only behaves like a particle

(e) Light behaves only like a wave, but matter behaves like a wave and like a particle

(f) Light behaves like a wave and like a particle, and matter behaves like a wave and like a particle
as well

7. Fill in each of the following blanks.

(a) de Broglie used the equations and to derive an equation for

the wavelength of a matter wave.

(b) Scientists first saw matter waves by looking for them in . This was a good

idea, because are small enough to have matter waves that can be observed in a

laboratory.

8. What is the wavelength of a 5.0 kg bowling ball that rolls down the lane at 2.0 m/s?

9. If you walk through a door at 1.0 m/s, and you weight 120 lbs (or 54 kg), what is your wavelength?
(This is also approximately the width of the door that would cause your body to diffract.)

10. A car has a mass of 1250 kg. If the car's wavelength is 2.41 x 10~ 38 m, at what speed is the car
traveling?

11. A bobsled sliding down the run at 14.8 m/s has a wavelength of 1.79 x 10~ 37 m. What is the total
mass of the bobsled?

Further Reading / Supplemental Links

• http : //www . stockcarroadrace . com/compete2008 . html

• http : //vergil . chemistry . gatech . edu/notes/quantrev/node6 . html

• http : //www . Colorado . edu/physics/2000/quantumzone/debroglie . html

• http : //www . launc . tased . edu . au/ONLINE/SCIENCES/physics/debroglie . html

• http://my.morningside.edu/slaven/Physics/uncertainty/uncertainty3.html

• http : //physics . about . com/od/lightoptics/a/waveparticle . htm

• http : //nobelprize . org/nobel_prizes/physics/laureates/1929/broglie-bio . html
www.ckl2.org 190

Vocabulary

wave-particle duality of matter Matter exhibits both particle-like and wave-like properties.

6.2 Schrodinger's Wave Functions

Lesson Objectives

• Distinguish between traveling and standing waves.

• Explain why electrons form standing waves, and what this means in terms of their energies.

• Define an electron wave function and electron density and relate these terms to the probability of
finding an electron at any point in space.

Introduction

In the last lesson, you learned that electrons and, in fact, all objects with mass, have wave-like properties.
It might be tempting to visualize matter waves as being just like ocean waves, or waves in a puddle, but it
turns out that matter waves are special. Unlike ocean waves or puddle waves, matter waves are 'trapped'
in space and, as a result, can never die out, escape, or disappear. If you think carefully, you'll realize that
this isn't true of most other waves with which you are familiar. You can form waves in a puddle by stirring
the puddle with a stick. When you do, what you'll notice is that the waves you create actually move from
your stick out to the edge of the puddle, where they disappear. As long as you disturb the puddle with
your stick, the puddle will have waves in it. But as soon as you leave the puddle alone, the surface of the
puddle will become as calm as glass. Matter waves aren't like that. Unlike puddle waves, which eventually
die out as they escape from the puddle, matter waves never do, because matter waves don't move. As a
result, they are forever trapped in the matter that holds them. We'll talk more about these special matter
waves in the next section.

An Electron is Described as a Standing Wave

Most of the waves that you're probably familiar with are known as traveling waves, because they travel
or move. When you're sitting on your surfboard, trying to catch a good wave, you'll often look out to
sea in the hopes of spotting a 'big one' (Figure 6.3). When you finally do, you know that even though
the big wave may be quite a distance off, it will eventually arrive at your surfboard and carry you in to
shore. This, of course, implies that ocean waves are traveling waves because they actually move through
the water. Similarly, if you're in the stands watching the Oakland A's play ball, you might find yourself
jumping up and cheering as 'the wave' passes through the stadium (Figure 6.4). Again, this is an example
of a traveling wave, because it moves from fans at one end of the stadium to fans at the other. There are,
however, special waves that stay in one spot. Scientists call these waves standing waves.

In an earlier part of this text, a wave was described in which a rope was tied to a tree and a person jerked
the other end of the rope up and down to create a wave in the rope. When a wave travels down a rope and
encounters an immoveable boundary (like a tree) , the wave reflects off the boundary and travels back up
the rope. This causes interference to occur between the wave traveling toward the tree and the reflected
wave traveling back toward the person. If the person adjusts the rhythm of their hand just right, they can
arrange for the crests and troughs of the wave moving toward the tree to exactly coincide with the crests
and troughs of the reflected wave. When this occurs, the apparent horizontal motion of the wave ceases
and the wave appears to "stand" in the same place in the rope. This is called a standing wave. In such

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Figure 6.3: When you're surfing, you wait for a good wave. Surfing is possible because ocean waves are
traveling waves. In other words, ocean waves actually move through the water.

Figure 6.4: Some fans getting ready for 'the wave' as it passes through the stadium.

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192

a case, the crests and troughs will remain in the same places and nodes will appear between the crests and
troughs where the rope does not appear to move at all.

Figure 6.5

In the standing wave shown in Figure 6.5, the positions of the crests and troughs remain in the same
positions. The crests and troughs appear to exchange places above and below the center line of the rope.
The flat places where the rope crosses the center axis line are called nodes (positions of zero displacement).
These nodal positions do not change.

Traveling waves appear to travel, and standing waves appear to stand still.

Even though standing waves don't move themselves, they are actually composed of traveling waves that
do. Standing waves form when two traveling waves traveling in opposite directions at the same speed
combine or run into each other. In today's lab, you'll learn how to create standing waves in a jump rope by
feeding traveling waves into the jump rope from opposite directions. Even though a standing wave doesn't
move, it can still 'die out'. As soon as the traveling waves that form a standing wave disappear, so does
the standing wave itself. You'll see this first hand in the jump rope experiment. When you stop flicking
the jump rope, the jump rope slackens and the standing waves are gone.

Why, then, are standing waves often associated with 'trapped' waves, or waves that never die out? The
connection between standing waves and trapped waves isn't a misconception or a misunderstanding. It
turns out that standing waves almost always form when traveling waves are 'trapped' in a small region
of space. Imagine what would happen if you took a whole train of traveling waves, locked them up and
threw them into jail. Those traveling waves would probably go crazy running around the jail cell trying
to escape. No matter how hard they tried, though, they'd always end up hitting the jail cell walls. As a
result, the poor waves would bounce back and forth and back and forth from one end of the jail cell to the
other. Now, if there were several traveling waves trapped in the same jail cell at the same time, one set
of waves would end up bouncing off of the left wall, at the same time (and speed) as another set of waves
was bouncing off of the right wall. This, of course, is exactly what's required to set up a 'standing wave'
(two waves traveling in opposite directions at the same speed).

The electron waves that you learned about in the last lesson form standing waves as a result of being
trapped inside the atom. What do you think might imprison an electron wave inside an atom? The
answer, of course, is that electrons are trapped because they are strongly attracted to the protons in the
nucleus. Using the laws of physics to describe the forces of attraction between electrons and protons,
scientists can figure out the size and shape of any electron's jail cell. Amazingly, by knowing the size and
shape of an electron's jail cell, scientists can tell you what a particular electron standing wave will look
like.

Frequently, rather than using words to describe an electron standing wave, scientists use what's known as an
electron wave function. Wave functions for electrons, first developed by a man named Erwin Schrodinger,

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are mathematical expressions that describe the magnitude or 'height' of an electron standing wave at every
point in space. Now, let's discuss electron energy, which is another important electron property that can
be explained and predicted by electron standing waves and their associated wave functions.

Each Wave Function has an Allowed Energy Value

Electrons form standing waves whenever they're trapped inside an atom, and thus in order to understand
and predict electron behavior, it's important to understand electron standing waves. One of the most
important properties that electron standing waves can help to predict is electron energy. The energy of
an electron in any atom depends on the size and shape of the electron's standing wave when it's trapped
inside that atom. As a result, scientists can use the wave function, or the mathematical description of an
electron's standing wave, to figure out how much energy that electron has.

While wave functions are helpful in predicting the amount of energy an electron has, they are also helpful
in predicting the amount of energy an electron is allowed to have. In any confined space, like a box, a jail
cell, or an atom, only certain standing waves are possible. Why? In order to exist, a standing wave must
begin at one side of the box and end at the other. Waves that either don't begin where the box begins, or
don't end where the box ends aren't allowed. Figure 6.6 shows several allowed standing waves and several
forbidden standing waves. Notice that if the wave doesn't 'fit' perfectly inside the box, it isn't allowed.

allowed standing waves

forbidden standing waves

KXXX) KXXXD

Figure 6.6: The standing waves on the left hand side of the figure are allowed to form in the brown box,
because they fit perfectly inside the box. In contrast, the waves on the right hand side of the figure cannot
form in the brown box because they do not fit perfectly inside the box.

Now here's the really strange thing about describing electrons as standing waves. Since only certain
standing waves will fit perfectly inside an atom, electrons trapped in that atom can only have certain
electron wave functions with certain electron energies. In other words, the standing wave picture accounts
for the fact that some energy values are ' allowed 1 (energy values associated with standing waves that 'fit'
perfectly inside the atom) while others are ' forbidden' energy values (energy values associated with standing
waves that do not 'fit' perfectly inside the atom). That's exactly what Bohr said when he developed his
model to explain atomic spectra! Bohr said that electrons could exist at specific ''allowed'' energy levels, but
that they couldn't exist between those energy levels. Bohr, however, did not have an explanation for why
only certain energy levels were allowed. Remarkably, the standing wave description of electrons predicts

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quantized electron energies just like the Bohr model!

When we represent electrons inside an atom, quantum mechanics requires that the wave must "fit" inside
the atom so that the wave meets itself with no overlap; that is, the "electron wave" inside the atom must
be a standing wave. If the wave is to be arranged in the form of a circle so that it attaches to itself, the
waves can only occur if there is a whole number of waves in the circle.

Figure 6.7

The standing wave on the left in Figure 6.7 exactly fits in the electron cloud and hence represents an
"allowed" energy level whereas the standing wave on the right does not fit in the electron cloud and therefore
is not an "allowed" energy level. There are only certain energies (frequencies) for which the wavelength fits
exactly to form a standing wave. These are the same energy levels the Bohr model suggested but NOW
there is a reason for why electrons may have ONLY these energies.

Max Born and Probability Patterns

There are very few scientists, if any, who can visualize the behavior of an electron during chemical bonding
or chemical reactions as standing waves. When chemists are asked to describe the behavior of an electron
during a chemical change, they do not describe the mathematical equations of quantum mechanics nor do
they discuss standing waves. The behavior of electrons in chemical reactions is best understood in terms
of a particle.

Erwin Schrodinger's wave equation for matter waves is similar to known equations for other wave motions
in nature. The equation describes how a wave associated with an electron varies in space as the electron
moves under various forces. Schrodinger worked out the solutions of his equation for the hydrogen atom and
the results agreed with the Bohr values for the energy levels of these atoms. Furthermore, the equation
could be applied to more complicated atoms. It was found that Schrodinger's equation gave a correct
description of an electron's behavior in almost all cases. In spite of the overwhelming success of the wave
equation in describing electron energies, the very meaning of the waves was vague.

A physicist named Max Born was able to attach some physical significance to the mathematics of quantum
mechanics. Born used data from Schrodinger's equation to show the probability of finding the electron, as
a particle, at the point in space for which Schrodinger's equation was solved. Born's ideas allowed chemists
to visualize the results of the wave equation as probability patterns for electron positions.

Suppose we had a camera with such a fast shutter speed that it could take a photo of an electron in an
electron cloud and show it frozen in position. We could then take a thousand pictures of this electron at
different times and find it in many different positions in the atom. We could then plot all the different
electron positions on one picture.

Figure 6.8 shows the result of plotting many different positions of a single electron in the electron cloud

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^ '&*££*#&>■. \

... %

■%-S'

Figure 6.8: The probability pattern for a single electron atom.

of a hydrogen atom. One way of looking at this picture is as an indication of the probability of where you
are likely to find the electron in this atom. You must recognize, of course, that the dots are not electrons;
this atom has only one electron. The dots are positions where the electron can be found at different times.
From this picture, it is clear that the electron spends more time near the nucleus than it does far away. As
you move away from the nucleus, the probability of finding the electron becomes less and less. It is also
important to note that there is no boundary for this electron cloud. That is, there is no distance from the
nucleus where the probability becomes zero.

For much of the work we will be doing with atoms, it is convenient to have a boundary for the atom. Most
often, chemists choose some distance from the nucleus beyond which the probability of finding the electron
becomes very low and arbitrarily draw in a boundary for the atom. Frequently, the boundary is placed
such that 90-r or 95-f- of the probability for finding the electron is inside the boundary (Figure 6.9).

Figure 6.9: Artificial boundaries for an atom.

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196

Most of the time, we will be looking at drawings of atoms that show an outside boundary for the electron
cloud. You should keep in mind, however, that the boundary is there for our convenience and there is
no actual boundary on an atom; that is, the probability of finding the electron never becomes zero. This
probability plot is very simple because it is for the first electron in an atom. As the atoms become more
complicated (more energy levels and more electrons), the probability plots also become more complicated.

Lesson Summary

• There are two types of waves - traveling waves that move from one place to another, and standing
waves that are stationary. Standing waves are formed when two traveling waves traveling in opposite
directions at the same speed combine. Electrons in atoms form standing waves because they are
trapped by the attractive forces that exist between their negative charges and the positive charges
on the protons in the atom's nucleus. These attractive forces determine the shape and size of the
electron's standing wave.

• Mathematical expressions called wave functions are used to describe an electrons standing wave in
an atom. The energy of an electron in any atom depends on the size and shape of the electron's
standing wave. The wave function can be used to determine the energy of an electron when it is
trapped inside an atom.

• Electrons in atoms are only allowed to have certain energy levels (ie - those which correspond to
standing waves that 'fit perfectly' inside the atom). All other electron energies are forbidden. The
probability patterns for electrons (electron density) show the probability of finding the electron at a
given point.

Review Questions

1. Choose the correct word in each of the following statements.

(a) The (more/less) electron density at a given location within the atom the more likely you are to
find the electron there.

(b) If there is no electron density at a particular point in space, there is (no/a high) chance of
finding the electron there.

(c) The higher the probability of finding an electron in a certain spot, the (more/less) electron
density there will be at that spot.

2. The hydrogen ion, H + has no electrons. What is the total amount of electron density in a hydrogen
atom?

3. Decide whether each of the following statements is true or false.

(a) Only certain electron standing waves are allowed in any particular atom.

(b) Only certain electron energies are allowed in any particular atom

4. The name for the mathematical expression used to describe an electron standing wave is -

5. Choose the correct statement.

(a) Einstein first developed the method of describing electron standing waves with wave functions

(b) Planck first developed the method of describing electron standing waves with wave functions

(d) Schrodinger first developed the method of describing electron standing waves with wave functions

6. Circle all of the statements below which are correct.

(a) The wave function description of electrons predicts that electrons orbit the nucleus just like
planets orbit the sun.

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(b) The wave function description of electrons predicts that electron energies are quantized

7. Fill in the blanks.

(a) Since only certain values are allowed for the energy of an electron in an atom, we say that
electron energies are .

(b) Allowed electron energies correspond to that fit perfectly in the atom.

8. Forbidden electron energies correspond to electron standing waves that in the atom.

Further Reading / Supplemental Links

• http://www. chemistry.ohio-state.edu/betha/qm

• http : //en . wikipedia . org/wiki/Many-worlds_interpretat ion

• http : //plato . Stanford . edu/entries/qm-manyworlds

• http://frombob.to/many.html

• http://en.wikipedia.org

Vocabulary

traveling waves Waves that travel, or move.

standing waves Waves that do not travel, or move. They are formed when two traveling waves, moving
in opposite directions at the same speed run into each other and combine.

electron wave function A mathematical expression to describe the magnitude, or 'height' of an electron
standing wave at every point in space.

electron density The square of the wave function for the electron, it is related to the probability of
finding an electron at a particular point in space.

6.3 Heisenberg's Contribution

Lesson Objectives

• Define the Heisenberg Uncertainty Principle.

• Explain what the Heisenberg Uncertainty Principle means in terms of the position and momentum
of an electron.

• Explain why the Heisenberg Uncertainty Principle helps to justify the fact that a wave function can
only predict the probable location of an electron and not its exact location.

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Introduction

If you think back to the last lesson, you'll remember that scientists had a lot of difficulty understanding
electron density and the wave function in terms of the wave properties and the particle properties of the
electron. Max Born found a way to use the electron wave function to calculate electron density, and that
electron density is actually equal to the probability of finding an electron at any point in space. This,
however, leads to the question, why can't scientists predict where the electron will be with certainty. Why
can they only predict the probability of finding an electron at any given point in space? Is there something
wrong with the theory? Is it possible to improve the theory so that scientists can predict exactly where an
electron is and where it's going?

Heisenberg Proposed the Uncertainty Principle for Behavior of
Electrons

The Uncertainty Principle

When scientists first suggested that the wave function was related to the probability of an electron being at
a specific point in space, it raised a lot of questions. Most importantly, scientists wondered why the wave
function could only predict the probability of finding an electron at a given location, and not the exact
location where the electron actually was. Some scientists suggested that the wave function couldn't make
exact predictions because it wasn't complete. They believed that the wave function was actually missing
information that was necessary to describe electron behavior with 'certainty'.

Some early scientists thought that perhaps the wave function could only predict the probability of an
electron being at a given location because the wave function was missing information. Many scientists
spent time looking for 'hidden variables' that controlled electron behavior just like the spin controls the
movement of a baseball. The assumption was that if the correct 'hidden variables' could be found and
included in the wave function, then the exact movement, and the exact location of an electron could be
predicted, just as easily as we can predict the movement of larger objects like baseballs, cars and planets.

Everything changed, however, in 1926 when a man named Werner Heisenberg (Figure 6.10) proposed
what's known as the Heisenberg Uncertainty Principle. According to the Heisenberg Uncertainty
Principle, it is impossible to measure certain properties, like momentum (speed multiplied by mass) and
position at the same time without introducing uncertainty into the measurement. Of course, if you can't
make accurate measurements, you can't make accurate prediction either.

What prevents scientists from making accurate predictions about small objects like atoms and electrons? Is
it that the machinery used to make the measurements is simply not good enough? Could scientists design
better machines and better measurements and then be able to predict electron behavior with certainty?
According to Werner Heisenberg, the answer is 'no - when it comes to small objects, scientists will never
be able to make accurate measurements and accurate predictions, no matter how good their machinery
is'. If you find that statement strange, you're not alone. Many scientists, even today, are bothered by the
Heisenberg Uncertainty Principle - it seems as if, with improved machines, we should be able to make better
measurements and thus better predictions! To some extent, that's true. Better machines can help us to
make better measurements and better predictions, but according to the Heisenberg Uncertainty Principle,
there is a fundamental limit to how much we can know and how accurately we can know it. It's as if there
is 'something' in the universe which prevents us from being able to make absolutely, one hundred percent
accurate measurements and, as a result, we will always be plagued by some uncertainty. For large objects,
like baseballs and planets, the uncertainty is just too small to matter, but for tiny things, like atoms and
electrons, the uncertainty becomes important.

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Figure 6.10: Werner Heisenberg

Impossible to Fix Both the Position of an Electron and Its Momentum

The Heisenberg Uncertainty Principle actually applies to a lot of different measurements, but often, scien-
tists are concerned with two in particular - position and momentum. You probably know what position
means, but momentum is a term that you don't hear a lot in everyday life. Momentum, p, is the quantity
that you get when you multiply an object's mass by its speed (to be truly correct momentum is actually
mass times velocity, but as in Chapter 1, we won't worry about the difference between speed and velocity).

In terms of the position and momentum, the Heisenberg Uncertainty Principle is as follows:

There is a fundamental limit to just how precisely we can measure both the position and the momentum
of a particle at the same time.

So how does the Heisenberg Uncertainty Principle relate to the electron and all the problems scientists have
interpreting the electron wave function? Well, if you think about it logically, the Heisenberg Uncertainty
Principle basically means that it's impossible to predict both exactly what the electron will do or exactly
where the electron will be found. Suppose, for instance, that you know the electron's precise position, then
according to the Heisenberg Uncertainty Principle, you can't know its precise momentum as well. In other
words, when you know where the electron is, you don't know where it's going (since where it's
going is determined by the velocity component of its moment). Suppose, on the other hand, that you know
the electron's precise momentum. According to the Heisenberg Uncertainty Principle, you can't know its
precise position as well. In other words, when you know where the electron is going, you don't
know where it is.

Obviously, there is always some uncertainty when it comes to electrons. You either don't know where they
are, or else you don't know where they're going. As a result, any theory that claimed to predict exactly
where the electron was, or exactly which path it would take as it traveled around inside the atom would

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go against the Heisenberg Uncertainty Principle. Luckily, the wave function description doesn't claim to
predict the precise behavior of the electron. Instead, it only makes statements about the probability of
finding the electron at one place or another.

In other words, the wave function model is consistent with the Heisenberg Uncertainty Principle. Moreover,
the Heisenberg Uncertainty Principle suggests that the electron wave function equation is as complete as it
can possibly be. It may not entirely predict electron behavior, but that isn't because the model is wrong,
or faulty. It's because, in our universe, there is a limit to how accurately we can actually know what tiny
objects like electrons are doing.

The Problem of Making Very Small Measurements

When you use a thermometer to measure the temperature of a volume of water, you place the thermometer
into the water and leave it there until the water and the thermometer have adjusted to the same tempera-
ture. Almost all solids and liquids expand when they are heated and contract when they are cooled. Each
substance, however, expands and contracts by different amounts. In the case of the mercury and glass in a
thermometer, the mercury expands and contracts faster than the glass and so as a thermometer is heated,
the mercury expands faster that the glass tube and the mercury runs up the tube. The glass has been
marked (calibrated) for each temperature so that you can read the correct temperature from the markings
on the tube. In the process of measuring the temperature of hot water, the thermometer is placed in the
water and the thermometer absorbs heat from the water so that its temperature becomes the same as the
water and you can read the temperature of the water from the temperature of the thermometer.

You should see, at least theoretically, that when the thermometer absorbs heat from the water, the water is
cooled down; that is, the temperature of the water decreases because of the heat lost to the thermometer.
Therefore, the temperature you get when you measure the temperature is not the same temperature
of the water that was present before you introduced the thermometer. The act of measuring the
temperature of the water changed the temperature. When the volume of the water is reasonably
large, the magnitude of the change caused by introducing the thermometer is not significant so we don't
bother to consider it. What about if the volume of the water is very small? If the volume of water is
200 mL and the thermometer absorbs 20 Joules of heat, the introduction of the thermometer might change
the temperature of the water by approximately 0.03° C. Which is certainly not a significant change. But
what if the volume of water whose temperature we were measuring was only 2 mL? Introducing the same
thermometer into this small volume might change the temperature of the water 3°C, which would certainly
be a significant change. The point is that when we measure very small things, the act of making the
measurement may change what we are observing.

Consider the method that humans use to see objects. We arrange for photons (quanta) of light to strike
the object and we see the object by the directions, angles, and colors of the photons that bounce off the
object and strike us in the eye or other light measuring instrument. If only red photons bounce back, we
say the object is red. If no photons bounce back, we say there is no object present. Suppose for a moment
that humans were gigantic stone creatures and we used golf balls to "see" with. That is, we would fire off
golf balls at our surroundings and the balls would bounce off objects and come back and enter our eyes
so we could see the object. If this were true, we could see mountains successfully and large buildings and
trees . . . but could we see butterflies or small flowers? Obviously, the answer is no. The golf balls would
simply knock small objects out of the way and continue on . . . they would not bounce back to our eyes.

In the case of humans trying to look at electrons, the photons we use to see them with are of significant
energy compared to electrons and when the photons collide with the electrons, the motion and/or position
of the electron would be changed by the collision. Heisenberg's Uncertainty principle tells us we cannot
be sure of both the location of the electron and the motion (path) of an electron at the same time. As a
consequence, scientists had to give up the idea of knowing the path the electron follows inside

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an atom.

Lesson Summary

• The Heisenberg Uncertainty Principle states that it is impossible to measure certain pairs of properties
like momentum (mass multiplied by velocity) and position at the same time without introducing
uncertainty into one or both of the measurements. In other words, it is impossible to know both the
exact momentum and the exact position of a particle at the same time.

• The Heisenberg Uncertainty Principle suggests that the electron wave function is complete and that
it does not predict the exact behavior of an electron because it is actually impossible to do so.

• The uncertainty that Heisenberg spoke of is not due to the failure or inadequacies of the measuring
equipment, but rather a fundamental limit imposed by our universe.

Review Questions

1. What types of things in everyday life are impossible to predict with absolute certainty?

2. Why is it impossible to predict the future with absolute certainty?

3. Fill in the blank. According to the Uncertainty Principle, it is impossible to know

both an electron's and momentum at the same time.

4. Decide whether each of the following statements is true or false:

(a) According to the Heisenberg Uncertainty Principle, we will eventually be able to measure both
an electron's exact position and its exact location at the same time.

(b) The problem that we have when we try to measure an electron's position and its location at the
same time is that our measuring equipment is not as good as it could be.

(c) According to the Heisenberg Uncertainty Principle, we cannot know both the exact position and
the exact location of a car at the same time.

5. Circle the correct statement The Heisenberg Uncertainty Principle...

(a) applies only to very small objects like protons and electrons

(b) applies only to very big objects like cars and airplanes

Further Reading / Supplemental Links

• http : //zebu . uoregon . edu/~imamur a/208/ j an27/hup . html

• http : //www . aip . org/history/heisenberg/p08 . htm

• http : //scienceworld . wolfram . com/physics/UncertaintyPrinciple . html

• http : //plato . Stanford . edu/entries/qt-uncertainty

• http://en.wikipedia.org/wiki
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Vocabulary

momentum (p) The quantity you get when you multiply an object's mass by it's velocity (which as far
as you're concerned is the same as its speed).

Heisenberg's Uncertainty Principle Specific pairs of properties, such as momentum and position, are
impossible to measure simultaneously without introducing some uncertainty.

6.4 Quantum Numbers

Lesson objectives

• Explain the meaning of the principal quantum number, n.

• Explain the meaning of the azimuthal quantum number, I.

• Explain the meaning of the magnetic quantum number, m/.

Introduction

We've spent a lot of Chapter 6 talking about waves, and electron waves in particular. While most of us
know what normal water waves look like, very few people have an understanding of what electron waves
look like. In the last lesson, we talked about electron density, and how an electron wave could be thought
of as representing the thickness or thinness of the electron density 'fog' at any point in space within the
atom. We considered the probability pattern for the electron in a hydrogen atom. Now let's consider some
more complicated atoms.

Schrodinger's Equation Provides Three Integers Used to Define
the Energy States and Orbital for an Electron

You should remember that electrons form standing waves whenever they are trapped within an atom. You
should also remember that only certain standing waves are allowed in any confined space because only
certain standing waves can fit perfectly inside that space (remember, a perfect fit requires that a wave
begin and end where its box begins and ends). In a one-dimensional box, it's easy to picture all of the
possible waves that fit perfectly inside. In three dimensions, it's a little harder. Unfortunately when it
comes to electrons in an atom, there's an additional complication on top of the fact that electron waves are
three-dimensional! It turns out that the electrons in an atom aren't confined to nice square or rectangular
boxes. Instead, they're confined to spherical boxes (which should make sense, since atoms are, after all,
tiny little spheres). In other words, electron waves inside an atom must begin and end on the surface of
a sphere. You may have a really good imagination, and an amazing ability to picture objects in three-
dimensions, but for most people, trying to figure out what these spherical three-dimensional waves look
like can be quite a challenge.

Luckily, that's how electron wave functions can help. Electron wave functions basically describe the
possible shapes that electron waves can take. We won't actually worry about the wave functions. Instead,
we'll only worry about specific numbers, called quantum numbers. Quantum numbers are always part
of an electron wave function and are extremely important when it comes to determining the shape of a
probability pattern.

When electron wave functions were first developed by a man named Erwin Schrodinger (Figure 6.11), his
goal was to show how a wave-like description of the electron could be used to understand the behavior

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of an electron in a hydrogen atom. In order to do this, Schrodinger first defined the size and spherical
shape of the hydrogen atom itself. Schrodinger then assumed an electron trapped within a hydrogen atom
formed a standing wave that fit perfectly inside without 'spilling out' or 'doubling over' on itself (it turns
out that in circular or spherical boxes, misfit waves don't 'spill out' so much as they 'double over', as shown
in Figure 6.12. Finally, Schrodinger supposed that an electron wave had to be continuous (remember,
something that is continuous has no gaps, holes or jumps).

Figure 6.11: A photograph of Erwin Schrodinger

This wave fits perfectly
on the circle

This wave, however,
doubles over on itself

Figure 6.12: When waves fit perfectly on a circle (or a sphere) they meet up with themselves and form a
closed loop. When waves fit perfectly on a circle, though, they end up 'doubling over' on themselves.

Amazingly, what Schrodinger discovered was that in order to satisfy his basic assumptions about the
electron wave both fitting inside the atom and being continuous, certain quantities in the electron wave
function had to be 'whole number integers'. Whenever Schrodinger assigned a whole number to these
quantities, he ended up with a wave that fit perfectly inside the hydrogen atom. However, whenever he
assigned a decimal number to these quantities, he ended up with a wave function that either 'doubled
over' on itself, or was discontinuous! The quantities that had to be assigned whole numbers soon became
known as quantum numbers. In the wave function for a hydrogen electron there are always three quantum
numbers. The first quantum number, n, is called the principal quantum number, the second quantum
number, {, is called the azimuthal quantum number, and the third quantum number, mi, is called the

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magnetic quantum number. Together these three quantum numbers define the energy state and orbital of
an electron, but we'll talk more about exactly what that means in the next section.

n = Principal Quantum Number (Energy Level)

The first quantum number, known as the principal quantum number, is given the symbol n. In order to
describe a valid standing wave, n has to have integer values, but there's an additional restriction on n as
well. The value of n must be a positive integer value (n = 1, 2, 3, . . .). In other words, n can never equal a
negative integer. In fact, n can never even equal 0! The principal quantum number gives you two different
clues as to what an electron wave looks like. First, it tells you how the electron density spreads out as you
move away from the center of the atom. For electron waves with low principal numbers, like n = 1, the
electron density is very thick right in close to the center of the atom, but then becomes rapidly thinner
as you move out. In contrast, for electron waves with high principal quantum numbers, like n = 6, the
electron density isn't as thick near the center of the atom, but is spread quite a bit further out. In general,
the higher the principal quantum number, the further away from the nucleus you'll be able to detect a
significant amount of electron density (Figure 6.13).

When "n" is a small number, the
electron probability is closer to
the nucleus.

X-

When "n" is a larger number the
electron probability is spread out
further from the nucleus.

Figure 6.13: When the principal quantum number, , is small, most of the electron density is found close
to the nucleus of the atom. When the principal quantum number is large, though, the electron density
spreads further out.

Sometimes, you will hear people say that the principal quantum number determines the 'size' of the electron
wave function. When people say this, they don't actually mean the absolute or total 'size' of the electron
wave. What they mean is how big or small the electron wave appears to be based on where most (usually
about 90%) of the electron density is concentrated. In an electron wave with a low principal quantum
number, the electron density is mostly found close to the center of the atom. Even though there is a tiny
bit of electron density at distances far away from the atom's nucleus, there's so little that you can't really
tell it's there. As a result, electron waves with low principal quantum numbers appear small. On the
other hand, in an electron waves with a high principal quantum number, the electron density is much more
spread out, and thus much thicker at distances far from the center of the atom. Therefore, electron waves
with high principal quantum numbers appear big.

The principal quantum number also describes the total number of nodes that the electron wave contains.
What are nodes? Nodes are places where the electron wave has absolutely no amplitude, or 'height'. Take
a look at the one-dimensional waves in the figure below. Can you spot the nodes (portion a.)? Now take a
look at the two-dimensional waves (portion b.). Can you spot the nodes? A three-dimensional wave with
nodes is something like an onion. Think of how the onion has layers and how in between the different
layers there's always an empty space or a break. If the onion was a three-dimensional wave, the breaks in

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between the onion layers would be like the nodes. The higher the principal quantum number, n, the more
nodes an electron wave has.

A node is any place where a wave has zero amplitude, or zero height. Both a. and b. illustrate different
ways of representing waves. In a. the amplitude corresponds to the height of the yellow line above (or
below) the black axis. When the yellow wave crosses the black axis, the wave has zero amplitude and thus
a node. In b., the amplitude of the wave corresponds to the thickness of the blue cloud. When there is no
blue cloud, the wave has zero amplitude and thus a node.

a ' a node

AAy\AA

. a node

The principal quantum number is extremely important, not only because it tells you something about
the 'size' of an electron wave and the number of nodes in an electron wave, but also because it tells you
something about the energy of that wave. If you think back to Chapter 4, you'll remember that negative
electrons like to be close to the positive nucleus because the energy of an electron is lower the closer it
is to a positive charge. What does that mean in terms of the principal quantum number? It means that
an electron wave with a lower principal quantum number, and electron density centered close to
the nucleus of the atom, will have a lower energy, while an electron wave with a higher principal
quantum number, and electron density spread further away from the nucleus of the atom, will have a
higher energy.

Similarly, as the number of nodes in an electron wave increases, the energy of the wave increases as well.
Think about the jump rope experiment. Do you remember how you created standing waves in a jump
rope? Did it take more or less energy on your part to get multiple waves to form along the rope? It should
have taken you more energy to create more waves. That's because in the process of creating more waves,
you also created more nodes and nodes are always associated with an increase in energy. Again, let's take
a look at what his means in terms of the principal quantum number. An electron wave with a lower
principal quantum number has fewer nodes, and thus will also have a lower energy. On the other
hand, an electron wave with a higher principal quantum number has more nodes and thus will have
a higher energy.

Notice that as n increases, both the 'size' of the electron wave and the number of nodes it has increase as
well. As a result, the energy of an electron wave always increases with n.

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1. The bigger value of the n, the higher the energy.

2. The smaller the value of n, the lower the energy.

Since the principal quantum number determines the energy of a particular electron wave, n is often thought
of as referring to the 'energy level' of the electron. The term 'energy level' actually comes from Bohr's old
solar system model of the atom. Thanks to Schrodinger, and his wave equation, though, we now know that
the energy level doesn't correspond to a particular orbit around the nucleus, but rather, to a particular
electron wave adopted by the electron when it becomes trapped inside the atom.

The Angular Momentum Quantum Number (Sub-levels, s, p, d,

The second quantum number, known as the azimuthal quantum number, is given the symbol I. While
the principal quantum number told you about the 'size' of an electron wave and the number of nodes in
an electron wave, the azimuthal quantum number tells you more about the 'shape' of an electron wave.
In other words, the shape that the electron wave appears to have as a result of its electron density being
'thicker' in one place than it is in another. You might be tempted to think that the shape of an electron
wave is always spherical because the atom itself is spherical. It turns out, however, that while there are
spherical electron waves, there are also waves that look like dumb-bells and waves that look like butterflies,
and waves that look so crazy it's almost impossible to describe them! Some of the different possible shapes
for electron waves are shown in the figure below.

You might wonder about the various balloon-like shapes in the figure above. One difficulty with using
drawings to represent electron waves is that the electron density itself is actually spread over a huge region
of space surrounding the center of the atom. As you learned earlier, though, for many electron waves,
almost all of the electron density is in close to the nucleus of the atom, with only a tiny bit of electron
density further out. Over the years, scientists have developed a standard way of drawing electron waves.
Rather than trying to account for all of the electron density in an electron wave, scientists usually just
draw 'balloons' around the region of space that contains about 90% of the electron wave's total electron
density. The figure below shows how scientists convert a cloud of electron density into a balloon. Even
though there is a little bit of electron density outside of the cartoon balloons (and thus a small probability
of finding the electron outside of its balloon), most electron behavior can be understood by ignoring the
tiny bit of electron density that the cartoon balloon doesn't capture.

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the electron density in an electron wave

90% of the electron density is inside the
two lobes drawn on the electron wave

c*o

these balloons represent the region of space
which contains 90% of the electron density
for this particular electron wave

The exact shape of the cartoon balloon representing an electron wave is determined by the value of t. In
other words, the dumb-bell shaped balloon in the image above has one value of £, while the butterfly-
shaped balloon has another value of t. Of course, scientists would get pretty tired of saying things like
'the dumb-bell shaped wave,' or 'the butterfly shaped wave,' so instead, they name the different waves
using letters from the alphabet. The most common shapes for waves are called s,p,d and /. (You would
probably find it a lot easier to remember what the different waves looked like if scientists had given them
nice descriptive names, like the 'dumb-bell wave' or the 'butterfly wave,' but boring names like s,p,d and
/ turn out to be much more convenient).

In the next lesson, we'll take a closer look at exactly what some of the common wave shapes look like.
First, though, we have to consider the relationship between n and I. Remember that a wave function for
an electron always has three quantum numbers. In order to fully describe an electron wave, then, you have
to know the values of all three of these numbers (n, I and mi). Now you might think that as long as n,
I and nil are all integers, the wave that they describe will be a perfectly good electron wave. But that's
NOT the way it works. It turns out that for a particular value of n, only certain values of I are allowed.

Warning! For a particular value of n, only certain values of I are allowed.

For wave functions that actually describe electrons in an atom, £ is always no less than 0, but also no more
than n — 1(£ = 0,1,2, ...n— 1). The following examples should help to clarify the restriction on I.

Example 1:

What are the allowable values of I for an electron wave with n = 3?
n = 3

1 . Find the minimum value of I.

The minimum value of t is always 0.

2. Find the maximum value of €.

The maximum value of I is always 1-1
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maximum t = n — 1
maximum t = 3 — 1
maximum I = 2

3. List a^ of the integers (no decimals!) starting from the minimum value of (, and ending with the
maximum value of I.

£ = 0,1,2

Example 2:

What are the allowable values of I for an electron wave with n = 1
«= 1

1 . Find the minimum value of (.

The minimum value of I is always 0.

2. Find the maximum value of t.

The maximum value of I is always 1-1
maximum i = 1 — 1
maximum i =

3. List a^ of the integers (no decimals!) starting from the minimum value of I and ending with the
maximum value of t.

£=0

Often, scientists will refer the different values of t as electron sublevels. In a hydrogen atom with one
electron, the value of t has no effect on the energy of the electron. However, in an atom with more than
one electron, the value of I does have a small effect on the electron's energy. In other words, the principle
quantum number, n, always determines the overall energy level of an electron, but that energy level is
actually divided into multiple sublevels based on the value of t. All of these sublevels have equal energy
in a hydrogen atom, because the hydrogen atom only has one electron. For atoms with more than one
electron, though, the different sublevels split apart, and some of them end up having more energy than
others.

ml =Magnetic Quantum Number (Identifies Orbital)

The third, and final quantum number, known as the magnetic quantum number, is given the symbol m\.
Remember, the principal quantum number told you about the 'size' of an electron wave and the number
of nodes in an electron wave, while the azimuthal quantum number told you more about the 'shape' of an
electron wave. The magnetic quantum number, though, gives you even more information about what the
electron wave looks like. The magnetic quantum number tells you how the electron wave is orientated in
space.

Orientation in space basically means where the electron wave points. Take a look at the two dumb-bell
shaped electron waves shown in Figure 6.14 (in the next lesson, you'll learn that these are actually p
orbitals). In the first electron wave, lobes of the wave point 'up-and-down' along the z-axis, while in the

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Figure 6.14: Two possible orientations for an electron wave with a dumb-bell shape.

second electron wave, the lobes of the wave point 'in-and-out' along the x-axis. These two electron waves
have different orientations in space, and thus different values for the quantum number m/.

Now compare Figure 6.14 with Figure 6.15. In both figures, the electron wave has a different orientation
in space, as indicated by the red arrow. But what do you notice about the orientations of the electron
wave in Figure 6.15? They look the same, don't they? Obviously, orientation doesn't matter for the
spherical electron wave. In other words, because the spherical electron wave looks the same no matter how
you rotate it, there's really just one orientation. So how many different mi values should an electron wave
with a spherical shape have (remember, mi values are used to describe orientation)? Clearly, an electron
wave with a spherical shape should only have one mi value, because it only has one possible orientation.
What about an orbital with a dumb-bell shape? Should it have a single mi value, or should it have several
different m/ values? Well, since different orientations of the dumb-bell shaped wave actually look different,
you'd probably expect the dumb-bell shaped wave to have several different m/ values, one for each possible
orientation. In fact, the dumb-bell shaped wave actually has three possible values for mi, and so it has
three possible orientations. (Don't worry about why there are exactly three orientations for the dumb-bell
shaped wave. We'll talk a little bit about that in the next lesson, but the full explanation requires a lot of
math that you'll learn about if you decide to study quantum physics or quantum chemistry). Hopefully,
by comparing Figures 8 and 9, you should be convinced that the shape of the wave (which depends on () is
important when it comes to determining the number of possible orientations (or the number of possible m/
values). It shouldn't be surprising, then, that for any given value of t, only certain mi values are allowed if
you want to end up with a wave function that makes sense and actually describes electrons in a hydrogen
atom.

Warning!

For a particular value of C, only certain values of m; are allowed.

The rule for m/ is that for any value of (, mi can be any integer starting from — I and ending at +€.
mi = -€, . . . , +i). The following examples should help to clarify the restriction on m/.

Example 3:

What are the allowable values of mi for an electron wave with t = 2?

1. Find the minimum value of m/.

The minimum, value of mi is always —I.
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y x

Figure 6.15: Two possible orientations for an electron wave with a spherical shape. Notice that there isn't
a noticeable difference between these orientations.

minimum mi = -2

2. Find the maximum value of m/.

The maximum value of mi is always +£
maximum mi = +2

3. List all of the integers (no decimals!) starting from the minimum value of mi and ending with the
maximum value of m\

mi = -2,-1,0,1,2

Example 4:

What are the allowable values of m; for an electron wave with I = 0?
£ =

1 . Find the minimum value of mi

The minimum value of mi is always —I.
minimum m/ =

2. Find the maximum value of m/.

The maximum value of mi is always +t
maximum mi =

3. List all of the integers (no decimals!) starting from the minimum value of mi and ending with the
maximum value of mi

mi =

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Now that we've talked about all three of the different quantum numbers, you should have a good under-
standing of how different electron waves can be described. You can describe the 'size' of an electron wave,
and the number of nodes in an electron wave using the principal quantum number, n. You can describe
the shape of an electron wave using the azimuthal quantum number, I. Finally, you can describe the
orientation of the electron wave using the magnetic quantum number, m/. Since you know how to describe
a general electron wave, it's time to look at several examples of specific electron waves. That's what we'll
talk about in the next lesson.

Lesson Summary

• Schrodinger discovered that in order for a wave function to describe a standing wave that was con-
tinuous, and that didn't 'doubled back' on itself, certain quantities in his wave function had to have
integer values.

The quantities in the wave function which must have integer values are known as quantum numbers.
In the wave function for hydrogen, there are three quantum numbers. They are called the principal
quantum number (n), the azimuthal quantum number (■£), and the magnetic quantum number (mi).
The principal quantum number can only have positive integer values, (n = 1,2, 3 . . .).
The principal quantum number determines how far the bulk of the electron density extends from the
center of the atom. The higher the value of n, the further away from the nucleus you will be able to
detect a significant amount of electron density.

The principal quantum number also determines the number of nodes in an electron standing wave.
The higher the value of n, the more nodes there are in the electron wave.

The higher the principal quantum number, the greater the energy of the electron. Therefore the
principal quantum number is determines the energy level of the electron.

The azimuthal quantum number, (, determines the shape of the electron wave. The values of (, are
also called the electron sublevels. They are labeled with the letters s, p, d, f, g, h, etc.
For a wave function that actually describes an electron in an atom, t is always no less than zero, but
also no more than n — 1(1 = 0, 1,2 . . . n— 1).

In atoms with more than one electron, £ has a small effect on the electron's energy.
The magnetic quantum number, m;, determines how the electron wave is orientated in space. For
any given value of £, mi can be any integer from —I to + I (mi = —I, ■ ■■ , +€) .

Review Questions

1. Match each quantum number with the property that they describe.

(a) n - \. shape

(b) I - ii. orientation in space

2. A point in an electron wave where there is zero electron density is called a .

3. Choose the correct word in each of the following statements.

(a) The (higher /lower) the value of n, the more nodes there are in the electron standing wave.

(b) The (higher/lower) the value of n, the less energy the electron has.

4. Fill in the blank. For lower values of n, the electron density is typically found the

nucleus of the atom, while for higher values of n, the electron density is typically found -

the nucleus of the atom.

5. Circle all of the statements that make sense: Schrodinger discovered that certain quantities in the
electron wave equation had to be integers, because when they weren't, the wave equation described

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waves which...

(a) were discontinuous

(b) were too small

(d) were too short and fat

(e) 'doubled back' on themselves

6. What are the allowed values of t for an electron standing wave with n = 4?

7. How many values of t are possible for an electron standing wave with n = 9?

8. What are the allowed values of mi for an electron standing wave with t = 3?

9. How many different orientations are possible for an electron standing wave with I = 4?
10. What are the allowed values of mi for n = 21

Further Reading / Supplemental Links

• http : //www . Colorado . edu/physics/2000/elements_as_atoms/quantum_numbers . html

• http : //dbhs . wvusd . kl2 . ca . us/webdocs/Electrons/QuantumNumbers . html

• http : //www . wwnorton . com/college/chemistry/gilbert/concepts/chapter3/ch3_2 . htm

• http : //www . sparknotes . com/testprep/books/sat2/chemistry/chapter4sect ion4 . rhtml

Vocabulary

quantum numbers Integer numbers assigned to certain quantities in the electron wave function. Be-
cause electron standing waves must be continuous and must not 'double over' on themselves, quantum
numbers are restricted to integer values.

principal quantum number (n) Defines the energy level of the wave function for an electron, the size
of the electron's standing wave, and the number of nodes in that wave.

node A place where the electron wave has zero height. In other words, it is a place where there is no
electron density.

azimuthal quantum number {€) Defines the electron sublevel, and determines the shape of the electron
wave.

magnetic quantum number (m/) Determines the orientation of the electron standing wave in space.

6.5 Shapes of Atomic Orbitals

Learning Objectives

• Define an electron orbital.

• Be able to recognize s orbitals by their shape.

• Be able to recognize p orbitals by their shape.

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Introduction

In the last lesson, we learned how the principal quantum number determines the size of an electron wave
(and the number of nodes), £ determines the shape of an electron wave, and m; determines the orientation
of an electron wave. Now the effects of n are probably easy to visualize. For bigger values of n, the electron
wave gets bigger, and it ends up with more nodes. Similarly, the effects of mi are easy to visualize as
well. For different values of m/, the electron wave gets rotated into different orientations. In other words,
the electron wave points in different directions. What about the effects of t ? You know that £ tells you
something about the shape of the electron wave. You also know that certain waves are spherical while
others are dumb-bell shaped, or butterfly shaped or just plain crazy shaped! But how do you know which
value of £ corresponds to which shape?

Unless you have a lot of training in mathematics, and can understand the wave function, there's really no
way for you to predict what the shape of a wave with a particular value of £ will look like. Really, you just
have to know, or be told - so that's what we'll do in the next few sections. We'll tell you what an £ =
wave looks like, and what an £ = 1 wave looks like.

When the Azimuthal Quantum Number is 0, The Electron Oc-
cupies an s Orbital

In the last lesson, you learned that different electron wave shapes have different names, and that these
names are always letters of the alphabet like s,p,d and /. [These letters were chosen on the basis of
observations of line spectra. Certain lines were observed as a "sharp" series; others were classified as "
principal," "diffuse" or "fundamental series, thus s,p,d,f] These letters correspond to the shape of the
electron wave, or at least the shape that the electron wave appears to take as a result of its electron density
being thicker in one place than it is in the other. Now this can get a little confusing, but remember, electron
waves describe electron density, or electron 'fog' which comes from interpreting the wave-like properties
of the electron. Nevertheless, we always have to be careful that we don't forget about the particle-like
properties of the electron as well. As a result, once scientists know the shape of a particular electron wave,
they will often switch and start describing the electron as a particle again. To do this, they use the name
for the shape of the wave, but rather than saying 'the electron wave has an s shape', or 'the electron
wave has a p shape', they will say 'the electron is in an s orbital', or 'the electron is in a p orbital'.
This makes it sound as if the electron is a particle again, and as if the orbital is some sort of box that the
particle is confined to, or at least some sort of territory that the particle patrols.

This switching back and forth between the wave-like description of the electron and the particle-like de-
scription of the electron may seem confusing, or annoying, or just plain strange, but scientists do it to
remind themselves that electrons are both particles and waves. So exactly what is an orbital? Technically
speaking, an orbital is a wave function for an electron defined by the three quantum numbers, n, £ and
mi. What the wave function describes, though, is a region in space with a particular shape, where you are
likely to find an electron. In terms of waves, the orbital describes the region in space where the electron
density is very thick. In terms of particles, the orbital describes the region in space where there is a high
probability of finding the electron (which should make sense, because wherever the electron density is thick,
there is also a high probability of finding the electron).

So far we've decided that £ describes the shape of an orbital, which in turn describes where the electron
density is thick (and there is a high probability of finding the electron) and where the electron density
is thin (and there is a low probability of finding the electron). We're still, however, working towards a
description of what these waves with regions of thick and thin electron density actually look like. Let's
begin with the simplest wave shape. The simplest wave shape occurs for £ = 0. Whenever an electron wave
is described by the quantum number £ = 0, we say that the wave function describes what's known as an s

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wave and thus the electron is in an s orbital, s waves can be big or small, depending on the value of n. s
waves can also have a huge number of nodes, or no nodes at all, again depending on the value of n. What
all s waves have in common, though, is their shape. That's what we'll talk about in the next section.

s Sublevels are Spherically Shaped

All I = electron waves are s waves, or waves from the s sublevel, and they all describe electrons in s
orbitals. As suggested in the previous section, all electron waves from the s sublevel have the same overall
shape, regardless of the value of n, regardless of their size, and regardless of the number of nodes they
contain, s orbitals always correspond to spherical waves. The quantum numbers n = 1 and I = describe
a small spherical wave with no nodes, the quantum numbers n = 2 and I = describe a larger spherical
wave with a single node, and the quantum numbers n = 3 and I = describe an even larger spherical wave
with two nodes. These waves all look slightly different, as shown in Figure 6.16.

Figure 6.16: Various orbitals. All of these orbitals have , but they have different values for . The first
orbital has , and thus is small and has no nodes. The second orbital has , and thus is larger and has one
node. The third orbital has , and thus is even larger and has two nodes.

Nevertheless, they are all spherical, because they all have I = 0. Their shapes don't change - only their
sizes and the number of nodes that they contain.

Now if you think back to an earlier lesson, you might remember something special about the different
orientations of a spherical wave. Do you remember what happened when we rotated the spherical wave so
that it pointed in different directions? It ended up looking the same, didn't it! No matter which way you
rotate a sphere, it always looks the same. So how many different mi values do you expect for a spherical
wave? One, of course! Now that you know spherical waves all have I = 0, you can use your rules for mi
to figure out exactly how many different m/ are allowed. If you look back to Example 4 in the previous
lesson, you'll see that we actually did that calculation. It turned out that there was only one allowable
value for m/, and that was m/ = 0. In other words, there is only one orientation of a spherical wave. It all
makes sense!

So what does a spherical wave really mean? It means that your probability of finding an electron at any
particular distance from the center of the atom only depends on the distance, and not on the direction.
You can see this in Figure 6.17.

Notice how it doesn't matter which direction you move as you travel from the center of the atom out,
your probability of finding the electron is the same whether you head straight up, or straight down, or
straight to the right, or straight to the left! The fact that electron density, and the probability of finding
an electron is independent of direction is a special property of s orbitals, as you'll see shortly when we
begin discussing some of the other possible electron orbitals like those for £ = 1.

215

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1 L

Figure 6.17: Notice that the amount of electron density (here represented by the intensity of the blue color)
doesn't depend on direction. It does, however, depend on distance from the center of the atom.

When the Azimuthal Quantum Number is 1, then ml Can Only
Be -1, or +1

All I = 1 electron waves are p waves, or waves from the p sublevel, and they describe electrons in what are
known as p orbitals. Unlike s orbitals, p orbitals are not spherical, so they can have different orientations
in space. Now that you know all p orbitals have t = 1, you should be able to figure out exactly how many
different p orbital orientations exist by using your rules for m/. (m/ is the quantum number associated with
the orientation of a particular orbital). Let's figure it out.

Example 1:

How many different p orbital orientations are possible?

£= 1

From now on, whenever you're told an electron is in a p orbital, you're expected to know that electron has
the quantum number £ = 1

The question how many p orbital orientations are possible, but what it's really asking is how many different
mi values are allowed when ( = 1. We've already done this type of problem.

1. Find the minimum value of mi.

The minimum value of mi is always —I.
minimum mi = -1

2. Find the maximum value of mi.

The maximum value of mi is always +t
maximum mi = +1

3. List all of the integers (no decimals!) starting from the minimum value of mi and ending with the
maximum value of m/.

mi = -1,0,1

In this case m/ can equal -1,0 or 1, so there are a total of 3 allowed values for mi, and thus 3 possible p
orbital orientations.

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The p Orbitals are Often Described as Dumb-bell Shaped

Even though you know that there are three possible orientations for p orbitals, you can't really predict
their shape unless you know a lot more about mathematics, physics and wave functions. When scientists
use the wave function to draw the shape of an electron's p orbital, though, they always end up with is
something that looks a lot like a dumb-bell. Not only that, the three different p orbitals (one with mi = — 1,
another with mi = 0, and the third with mi = 1) turn out to be perpendicular to each other. In other
words, if one p orbital points along the x-axis, another p orbital points along the y-axis, and the third
points along the z-axis. Scientists typically label these three orbitals p x ,Py and p z respectively. Figure ??
shows each of the three p orbitals separately, and then all three together on the same atom.

Sometimes we get so caught up thinking about electron wave functions, and electron orbitals, that we
forget entirely about the atom itself. Remember, electron standing waves form because electrons get
trapped inside an atom by the positive charge on the atom's nucleus. As a result, s orbitals, and p orbitals
and even d and / orbitals always extend out from the atom's nucleus. Don't get so caught up in orbitals
that you forget where they are and why they exist.

As with s orbitals, p orbitals can be big or small, depending on the value of n, and they can also have
more or less nodes, also depending on the value of n. Notice, however, that unlike the s orbital, which can
have no nodes at all, p orbital always have at least one node. Take a look at the P orbital figure above
again. Can you spot the node in each of the p orbitals? Since all p orbitals have at least one node, there
are no p orbitals with n = 1. In fact, the first principal quantum number, n, for which p orbitals are
allowed is n = 2. Of course you could have figured that out for yourself, right? No? Well, here's a hint -
remember the rules for predicting which values of I are allowed for any given value of n. In the last lesson,
you learned that t must be no less than 0, but also, no greater than n— 1. For the n = 1 energy level, then,
the maximum allowed value for £ is

maximum I = n -1
maximum t = 1 — 1
maximum I =

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As a result, only s orbitals [i = 0) are allowed. For the n = 2 energy level, though, the maximum allowed
value for I is

maximum I = n — \
maximum t = 2 — 1
maximum t — 1

which means p orbitals {€ = 1) are allowed as well. So now you see that the restrictions on € are actually
there to make sure that all n = 1 wave functions have no nodes, all n = 2 wave functions have 1 node, all
n = 3 wave functions have 2 nodes, all... well, you get the picture.

One interesting property of p orbitals that is different from s orbitals is that the total amount of electron
density changes with both the distance from the center of the atom and the direction. Take a look at
Figure ??. Notice how the electron density is different depending on which direction you travel from the
center of the atom out. In the particular p orbital shown, the probability of finding the electron is greater
as you head straight up from the center of the atom than it is as you head straight to the left or to the right
of the atom. It turns out that this dependence on direction is very important when it comes to studying
how different atoms interact and form bonds. We'll talk more about that in a later chapter.

For p orbitals, the amount of electron density, and thus the probability of finding an electron, depends on
both the distance from the center of the atom and the direction. (Source: CK-12 Foundation. CC-BY-SA)

There is a high
probability of finding the
electron in this direction.

There is very low < Tnere is ver V low

probability of finding -^ |L ^- probability of finding

the electron in this tne electron in this

direction. direction.

There is a high
probability of finding the
electron in this direction.

The d Orbitals and f Orbitals are Not Easily Visualized

Did you notice how the p orbitals looked a lot fancier than the simple spherical s orbitals? Well, you can
imagine that if p orbitals with t = 1 are fancy, then d orbitals, with I = 2 are even fancier, and / orbitals,
with I = 3 are just plain crazy! Most people can visualize p orbitals, but d orbitals and / orbitals are
actually rather difficult to imagine. Most d-orbitals are butterfly shaped, although one has an unusual
shape that looks like a donut surrounding a Q-tip! The 5 possible d orbitals are shown in Figure ??

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(and you could have figured out that there were 5, right?). Don't worry too much about why one of the
d orbitals is different. Again, it takes a lot of complex math to understand where the different d orbital
pictures came from, and you won't have to worry about that unless you decide to go on and study quantum
chemistry at the university level. As for / orbitals, they are even hard to draw, not to mention the fact
that there are a total of 7 of them (Figure ??)! We won't be too concerned with d orbitals in this course,
although they do become very important if you want to study certain metals like those found in the center
of the periodic table. Similarly, / orbitals aren't all that important when it comes to common chemicals
like hydrogen, or oxygen, or even copper. They do become important, though when you want to study
some of the most important radioactive elements like uranium and plutonium!

The probability patterns for the d orbitals. (Source: CK-12 Foundation. CC-BY-SA)

x -y

The probability patterns for the f orbitals. (Source: CK-12 Foundation. CC-BY-SA)

rf A%^

219

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Information About the Quantum Theory
General Information

You should keep in mind that the theories scientists eventually accept are those that are in agreement with
observations on nature and offer explanations for those observations. Theories about science that are the
result of only mathematical development usually are not accepted until supportive observations are made.
Such was certainly the case with Einstein's theory of relativity.

In the case of quantum theory, even though it was developed primarily from observations on black-body
radiation and light emissions from atoms, its application is much broader. The primary tenet of quantum
theory is that energy comes in small packages called quanta. It doesn't say only light energy comes in
packages, it says all energy comes in packages. One of these packets of energy is called a quantum (plural
= quanta). In the specific case of light, a packet of energy can also be called a photon. Quantum theory
says that all energy is quantized (comes in little bundles). That means that even the kinetic energy of
a moving baseball must increase or decrease in steps. That is, a baseball may not have "any" possible
energy but rather must have stepwise increases or decreases in energy. This stepwise change in energy is
not possible to observe in baseballs, of course, because the change in the velocity of a baseball with the
increase or decrease of one quantum of energy would be much too small for us to detect. In order for us
to actually see the stepwise changes in energy of an object, the object must be extremely small so that a
change in energy of a single quantum would be measureable. Of course, such observations are made in the
gain or loss of energy by the electrons in the electron cloud of an atom when precise frequencies of light
(corresponding to individual photons) are absorbed or emitted when electrons change energy levels.

But, the stepwise change in energy is also detectable in the behavior of other very small objects. Molecules,
in addition to their linear motion, also rotate. This rotational motion requires very small levels of energy
and shows a stepwise increases and decreases in rotation. That is, certain molecules can be observed to
rotate at 3 revolutions per second or 6 revolutions per second, or 9 revolutions per second, but cannot be
observed to rotate at speeds in between those observed values. Molecules also have vibrational motion in
their chemical bonds. This vibrational motion is also observed to be stepwise in nature. These are just
some of the observations that are supportive of the quantum theory.

Quantum Numbers and Electron Arrangement

Quantum theory in general and specifically solutions to Schrodinger's equation gives us a great deal of
information about the arrangement of electrons in the electron clouds of atoms. When we interpret this
information, it is quite apparent that it came from a mathematical development. Keep in mind, however,
that eventually you will see laboratory observations that have no explanation without quantum theory,
just as the spectra of elements were observed 100 years before anyone could explain why it occurred.

The principal quantum number, n, that comes from solutions to Schrodinger's equation indicates the major
energy level that contains that electron. These energy levels are numbered 1,2,3,... etc. (The energy levels
were originally named K,L,M,N,. . .etc, but later they were re-named 1,2,3,... and so on. Sometimes, in
older text material, you will see references to the K energy level.) Since Schrodinger's equation is a
theoretical statement, an infinite number of energy levels exist. However, when we place the electrons for
our known atoms into the energy level structure, the electrons from even the largest known atom all fit in
just 7 energy levels. There might be some theoretical discussion of an 8 or higher energy level but we
never actually use any above 7.

These energy levels also contain sub-levels. The sub-levels also have been named. The names of the sub-
levels are s, p, d, and /. Mathematically, the number of sub- levels that an energy level can have is equal to
the principal energy level number. That is, the 1 st energy level can have 1 sub-level, the 2 energy level

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can have 2 sub-levels, and so on. Obviously, if the number of sub-levels possible is equal to the energy level
number, then energy level 6 could theoretically have 6 sub-energy levels. Therefore, some people list the
sub-energy levels as s, p, d, f, g, h, i, . . . etc. Again, however, we find that our largest atom with the greatest
number of electrons never uses any sub-level beyond f. Therefore, it is often said that the largest number
of sub-levels used by atoms is 4.

The sub-energy levels have the probability patterns for electrons that you saw in the last section. The
s sub-energy level produces spherical probability patters, the p sub-levels produce the dumbbell shaped
probability patterns, the d sub-energy levels produce the butterfly probability patterns, and the / sub-levels
produce the probability patterns with no verbal description - but they do have a mathematical description.
The n quantum number identifies the sub-level that contains the electron. When I = 0, the electron is in
an s-orbital, when t = 1, the electron is in a p-orbital, when I = 2, the electron is in a J-orbital, and when
I = 3, the electron is in an /-orbital. For all our known atoms, I never gets larger than 3.

The sub-energy levels contain various spatial orientations of probability patterns called orbitals. This
word is a carryover from the Bohr theory where an orbital was a circular path the electron followed around
the nucleus but that is not the meaning of the word in modern chemistry. In modern chemistry, an orbital
is a specifically designated volume inside the electron cloud where the probability of finding the electron
is high. As mentioned earlier, the first energy level, n = 1, has only one sub-energy level because I must
equal and that sub-level is an s sub-level. The number of orbitals in the s sub-levels was calculated in
the last section. Sub-energy level s has a single orbital that is spherical in shape. Sub-energy level p has
three dumbbell shaped orbitals, while sub-energy level d has five orbitals, and sub-energy level / has seven
orbitals (Table 6.1).

Table 6.1: Summary of Energy Levels, Sub- Levels, and Orbitals

Energy Level Number Sub-Energy Levels Number of Orbitals

1 s 1

2 s 1

2 p 3

3 s 1
3 p 3

3 d 5

4 s 1
4 p 3
4 d 5

4 f 7

5 s 1
5 p 3
5 d 5
5 f 7

For actual atoms, all the energy levels after 4 are repetitions of 4. That is, they have the same sub-energy
levels and orbitals as energy level 4. In theory, of course, energy level 5 could have five sub-energy levels
and the fifth sub-energy level would be named g and have 9 orbitals. We don't normally discuss that
sub-energy level or any beyond it because they are never used. You might note the mathematical tone
in this data, in that the total number of orbitals in any energy level is equal to the energy level number
squared, n 2 . You probably also noticed that the number of orbitals in each succeeding sub-level increases
by 2.

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Lesson Summary

• An orbital is a wave function for an electron denned by the three quantum numbers, n, I and m/ .
Orbitals define regions in space where you are likely to find electrons.

• s orbitals {t = 0) are spherical shaped.

• p orbitals (t = 1) are dumb-bell shaped.

• The three possible p orbitals are always perpendicular to each other.

Review Questions

1. Fill in the blanks. When I = 0, the electron orbital is and when I = 1, the electron

orbital is shaped.

2. The n = 1 s orbital has nodes.

3. The n = 2 s orbital has nodes.

4. The n = 2 p orbital has nodes.

5. The n = 1 p orbital has nodes.

6. There are different p orbitals.

7. What energy level (or value of n) has s, p and d orbitals, but no / orbitals?

8. How many different d orbital orientations are there?

9. How many / orbital orientations are there?

10. How many different orbitals are there in the n = 3 energy level?

Vocabulary

orbital A wave function for an electron defined by all three quantum numbers, n, £, and mi. Orbitals
define regions in space where there is a high probability of finding the electron.

Image Sources

(2
(3
(4
(5
(6
(7
(8
(9
(10

(11
(12

Richard Parsons. . CC-BY-SA.

Richard Parsons. . CC-BY-SA.
Richard Parsons. . CC-BY-SA.
http : //en . wikipedia . org/wiki/Image : Surf er_above_the_wave . jpg. GNU-FDL.

Werner Heisenberg. Public Domain.
CK-12 Foundation. . CC-BY-SA.

http : //en . wikipedia . org/wiki/File : Conf ed-Cup_2005_-_Laolawelle . JPG. GNU-FDL.

www.ckl2.org 222

(13) A photograph of Erwin Schrodinger. Public Domain.

(14) .

(15) http://www.flickr.com/photos/davehogg/129247229/. CC-BY.

(16) .

(17) Louis de Broglie. Public Domain.

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Chapter 7

Electron Configurations for
Atoms

7.1 The Electron Spin Quantum Number

Lesson Objectives

• Explain what is meant by the spin quantum number, m s .

• Explain how the spin quantum number affects the number of electrons in an orbital.

• Explain the difference between diamagnetic atoms and paramagnetic atoms.

Introduction

In the Quantum Mechanics Model of the Atom chapter, you learned about quantum numbers. Remember,
quantum numbers actually come from the wave function. Even though you've never seen a mathematical
equation for a wave function (and you'd probably think it was pretty scary if you did!) you can still
understand what the different quantum numbers mean, because they all control different aspects of what
the electron standing wave looks like. The principal quantum number, n, determines the 'size' and the
number of nodes in the electron wave, the azimuthal quantum number, £, determines the 'shape' of the
electron wave, and the magnetic quantum number, mi, determines the 'orientation' of the electron wave.
Now suppose that you were told the exact size, the exact shape and the exact orientation of an electron
orbital. Could you draw that object without being told anything else? You should be able to - what could
you possibly need to know about an electron wave other than its size, shape and orientation? It seems
as if the first three quantum numbers tell us all the information necessary to picture what an electron's
probability pattern looks like - but it turns out that there's fourth quantum number! What! How could
there be another quantum number? What property could it possibly describe? There's nothing left after
size, shape, and orientation. That's exactly what scientists thought until they discovered spin.

m s = +1/2 or -1/2

In the early days, when scientists were just beginning to learn about quantum physics, most of what they
knew about quantization involved atomic spectra. Remember, in The Bohr Model of the Atom chapter, you
learned that atomic spectra were discontinuous, and that this led Bohr to propose the existence of quantized
energy levels. To explain these quantized energy levels, Bohr suggested that electrons in an atom were
restricted to specific orbits and that they moved around the nucleus in these orbits just like the planets

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move around the sun. It turned out that Bohr's model wasn't entirely correct because electrons don't
orbit the nucleus. In fact, based on what scientists know of the wave-particle duality and the Heisenberg
Uncertainty Principle, scientists now think that electrons can be anywhere inside the atom, although the
probability of finding them at any one particular location over another depends on the relative amount
of electron density at those locations. Even though Bohr's model had some problems, Bohr was right in
his prediction that electrons could only exist at specific 'allowed' energy levels, and that all other energies
were 'forbidden.'

In the Quantum Mechanics Model of the Atom chapter, you learned how the 'allowed' and 'forbidden'
energy levels could be explained in terms of electron standing waves. Only certain electron standing
waves will fit perfectly inside the atom without 'doubling over' on themselves, or containing discontinuous
jumps. Therefore, just like the Bohr Theory of the Atom, the wave function theory of the electron explains
the existence of discontinuous atomic spectra. Actually, wave functions themselves are pretty good at
predicting exactly what the atomic spectra of different atoms will look like (In other words, at which
wavelengths lines of light will appear - if you have forgotten about atomic spectra, go back and read The
Atomic Theory chapter!) Remember, Bohr's model could only predict the atomic spectrum of hydrogen.
With a little bit of work, though, scientists could get electron wave functions to predict many different
atomic spectra for many different types of atoms. It seemed perfect... but there was a still a slight problem.
Every so often when scientists looked carefully at atomic spectra, they would find two separate lines of
light at a wavelength where the wave function had only predicted one. The two separate lines were always
really close together, so if you didn't look carefully, you'd think it was just one line and you'd think that
the wave function was doing a great job. But the fact remained that frequently the wave function was
missing some of the finer details of the atomic spectra.

Now if you think back to what we discussed in The Bohr Model of the Atom chapter, you'll remember that
the lines of light in an atomic spectrum result when an electron 'falls' from one energy level to another
and releases its extra energy in the form of light. Even though we learned about this process in terms of
the Bohr model, the same principle applies to the wave function theory as well. Remember, in terms of
wave functions, the energy level of an electron is determined by the principal quantum number, n, while
the azimuthal quantum number, {, defines the sublevel and also affects electron energy in atoms with more
than one electron.

Electrons don't circle the nucleus like planets in a solar system as pictured in the figure below on the left.
Nevertheless, when they fall from a higher energy level electron standing wave to a lower energy electron
standing wave they release light in much the same way as shown in the figure below on the right.

When an electron begins at one energy (one value of n and I) and then falls to a lower energy (with a
different value of n and £), it releases its extra energy in the form of light. Therefore, if there are two very
closely spaced lines of light in an atomic spectrum, it must mean that there are two very closely spaced
energy states from which (or to which) an electron can 'fall.' In terms of predicting the finer details of an

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atomic spectrum, then, it's clear that the problem with the wave function was that in certain situations it
would only find one allowed energy state where there should have been two!

Figure 7.1: George Uhlenbeck (on the far left) and Samuel Goudsmit (on the far right), the two scientists
who first proposed the existence of the spin quantum number.

The story got even stranger when scientists put magnets around different atoms and looked at what
happened to their atomic spectra as a result. All of those pairs of closely spaced spectral lines that the
wave function couldn't predict actually split apart even further! In other words, magnetic fields affected
the two different energy levels differently. It made the low energy level drop even lower, and it made the
high energy level rise even higher. After the magnet experiments, scientists came to the realization that
their wave function description was not as complete as they had hoped. There was some other property
that had to be included - but what?

In 1925, two scientists by the names of Samuel Goudsmit and George Uhlenbeck (Figure 7.1) suggested
extending the wave function equation so that it included a fourth quantum number, called the 'spin'
quantum number. When any charged object (like, for example, a negatively charged electron) spins in
a magnetic field, its energy is determined by the direction in which it rotates. In other words, if the
object rotates clockwise, it will have a different amount of energy than it would have if it had rotated
counterclockwise. Goudsmit and Uhlenbeck argued that if the electron was spinning, then it was easy to
explain why those two closely spaced spectral lines split further apart in the presence of a magnetic field.

One line corresponded to the energy level for the electron spinning clockwise, and the other corresponded
to the energy level for the electron spinning counterclockwise. (If you're really clever, you might wonder
why these two energy levels were different and could be seen in atomic spectra even when there weren't
any external magnets around. It turns out that there are always tiny internal magnetic fields created by
matter itself. Usually, though, these internal magnetic fields are very small, so their effect is very small as
well.)

Pictured below are two negatively charged particles spinning in a magnetic field. The left particle will
have a different energy than the right particle, because they are spinning in different directions. (Source:
CK-12 Foundation. CC-BY-SA)

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negatively charged particle
spinning counterclockwise

negatively charged particle
spinning clockwise

magnetic field
* V V t V

■

By incorporating electron spin into the electron wave function, scientists found that the fourth quantum
number, also known as the spin quantum number, m s , could take on two different values - they were
m s = +1/2 and m s = -1/2. Notice that unlike n, I and mi, which can only be integers, m s can be half-
integers. The value of m s will never be some crazy decimal like m s = 0.943895, but it can be the decimal
m s = 0.5, because that corresponds to one-half, or a half-integer.

Now it's tempting to think of m s = +1/2 as the electron spinning in one direction, and m s = -1/2 as the
electron spinning in the other direction. When you do that, though, how are you picturing the electron?
You're picturing it as a particle, aren't you? And how do you explain the wave-like nature of an electron
in terms of rotation clockwise or counterclockwise? Well, you can't. So despite the fact that the idea for
including electron spin in the wave function came from picturing electrons as tiny little spinning objects,
scientists try not to make any direct comparisons between the spin quantum number and a particle rotating
clockwise or counterclockwise. Instead, scientists usually make vague statements like 'spin can only truly
be understood as a quantum property. There is no directly comparable macroscopic (large scale) property'.
Rather than saying anything about a clockwise rotation, or a counterclockwise rotation, scientists call
electrons with m s = +1/2 'spin-up' and electrons with m s = -1/2 'spin down'.

When Two Electrons Occupy the Same Orbital, They Will Have
Opposite Spins

Do you remember what an orbital is? An orbital is a wave function for an electron with a specific set of
quantum numbers, n, I, and ra/. For example, the numbers n = 1, £ = and m/ = define one particular
orbital, while the numbers n = 2, £ = 1 and mi = -1 define another orbital, and the numbers n = 2, I = 1
and mi = define a different orbital again. Orbitals are very important because any time you know the
values of the first three quantum numbers n, £ and mi, you know the region in space where there is a high
probability of finding the electron. In other words, if you know which orbital an electron is in, you know
exactly what its 'box', or 'territory' looks. The orbital is really just a description of where the electron
spends most of its time.

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For example, if you're told that the electron is in an orbital with n = 1, I = 0, mi = 0, you automatically
know that the electron is probably going to be found in a spherical shaped 'territory' that's pretty close
to the nucleus of the atom. On the other hand, if you're told that the electron is in an orbital with n = 5,
I = 1 and mi = you know that the electron is probably going to be found in a dumb-bell shaped 'territory'
that extends quite far from the nucleus of the atom, along the y axis (technically speaking mi = could
be the orbital along the x-axis, the orbital along the y-axis or the orbital along the z-axis, but whichever
axis you choose for the m; = orbital, the mi = 1 and mi = -1 orbitals will automatically point along the
other two axes).

The spin quantum number doesn't tell you anything about the region in space where you're likely to find
the electron. Therefore, if you want to know which orbital an electron is in, all you need to know is the
values of the first three quantum numbers - you don't need to know the value of the fourth. So why is the
value of the fourth quantum number important? Well, the spin quantum number determines whether or
not an electron is allowed to share an orbital with another electron that's already there. It turns out that
each region of the atom (or each orbital) can actually be shared by two electrons.

Two electrons can share their territory, or their orbital, provided each electron does something different.
This means that sharing an orbital is only possible for electrons which have different spin quantum numbers.
An electron that has a spin quantum number of m s = +1/2 can share an orbital with an electron that has a
spin quantum number of m s = -1/2. However, an electron that has a spin quantum number of m s = +1/2
cannot share an orbital with another electron that also has a spin quantum number of m s = +1/2. Similarly,
an electron that has a spin quantum number of m s = -1/2 cannot share an orbital with another electron
that has a spin quantum number of m s = -1/2.

When Electrons are Paired, They are Diamagnetic i.e. No Mag-
netic Attraction

In the last section, you learned that any time two electrons share the same orbital, their spin quantum
numbers have to be different. In other words, one of the electrons has to be 'spin-up', with m s = +1/2,
while the other electron is 'spin-down', with m s = -1/2. This is important when it comes to determining
the total spin in an electron orbital. Technically speaking, an electron's spin isn't exactly the same as its
spin quantum number, but the difference isn't important when it comes to figuring out whether or not two
electron spins cancel out. In order to decide whether or not electron spins cancel all you need to do is
add their spin quantum numbers together. If the total is 0, then the spins cancel each other out. If the
total is greater than 0, or less than 0, then the spins do not cancel each other out. Let's take a
look at two examples:

Example 1:

Electron A has m s = +1/2 while electron B has m s = -1/2. Do the spins of these two electrons cancel each
other out?

m s for A = +1/2

m s for B = -1/2

Total m s = m s for A + m s for B
Total m s = (+1/2) + (-1/2)
Total m s = 1/2-1/2
Total m s =

In this case, the spins of electron A and electron B do cancel each other out.
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Example

Electron A has m s

other out?

m s for A =

-1/2

m s for B =

-1/2

-1/2 while electron B has m s = -1/2. Do the spins of these two electrons cancel each

Total m s = m s for A + m s for B
Total m s = (-1/2) + (-1/2)
Total m s = -1/2 - 1/2
Total m. = -1

In this case, the spins of electron A and electron B do not cancel each other out.

The idea of 'canceling out' should make sense to you. If one spin is 'up,' and the other is 'down,' then the
'up' spin cancels the 'down' spin and there is no leftover spin at all. The same logic applies if we think
of spins as 'clockwise' and 'counterclockwise.' If one spin is 'clockwise' and the other is 'counterclockwise,'
then the two spin directions balance each other out and there is no leftover rotation at all. Notice what
all of this means in terms of electrons sharing an orbital. Since electrons in the same orbital always have
opposite values for their spin quantum numbers, m s , they will always end up canceling each other out! In
other words, there is no leftover (or 'net') spin in an orbital that contains two electrons. Whenever two
electrons are paired together in an orbital, we call them diamagnetic electrons. On the other hand, an
orbital containing only one electron will have a total spin equal to the spin of the electron that it contains.

Even though electron spin can only be truly understood using quantum physics, it does produces effects
that we can actually see in our everyday lives. Electron spin is very important in determining the magnetic
properties of an atom. If all of the electrons in an atom are paired up and share their orbital with another
electron, then the total spin in each orbital is zero and, by extension, the total spin in the entire atom is zero
as well! When this happens, we say that the atom is diamagnetic because it contains only diamagnetic
electrons. Diamagnetic atoms are NOT attracted to a magnetic field. In fact, diamagnetic atoms are
slightly repelled by magnetic fields.

Since diamagnetic atoms are slightly repelled by a magnetic field, it is actually possible to make certain
diamagnetic materials float! Below you can see a thin black sheet of pyrolytic graphite floating above the
gold magnets. (Source: http://en.wikipedia.Org/wiki/Image:Diamagnetic_graphite_levitation.
jpg. Public Domain)

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When an Electron is Unpaired, Paramagnetism will be Observed

If atoms that contain only paired electrons are slightly repelled by a magnetic field, what do you think
is true of atoms that contain unpaired electrons? If you guessed that atoms with unpaired electrons are
slightly attracted to a magnetic field, then you guessed right! Whenever electrons are alone in an orbital,
we call them paramagnetic electrons. Remember, if an electron is all by itself in an orbital, the orbital
has a 'net' spin, because the spin of the lone electron doesn't get canceled out. If even one orbital has a
'net' spin, the entire atom will have a 'net' spin as well. Therefore, we say an atom is paramagnetic
when it contains at least one paramagnetic electron. Notice that the definition of diamagnetism and
the definition of paramagnetism are subtly different. This can be confusing if you aren't careful. Be sure
to note: In order for an atom to be diamagnetic, all of its electrons must be paired up in orbitals. In order
for an atom to be paramagnetic, at least one of its electrons must be unpaired.

In other words, an atom could have 10 paired (diamagnetic) electrons, but as long as it also has 1 unpaired
(paramagnetic) electron, it's still considered a 'paramagnetic atom'. In order to be a 'diamagnetic atom',
the atom would have to have 10 paired (diamagnetic) electrons and no unpaired (paramagnetic) electrons.
Just as diamagnetic atoms are slightly repelled from a magnetic field, paramagnetic atoms are
slightly attracted to a magnetic field.

Lesson Summary

• If you only consider the first three quantum numbers, the wave function model for the electron will
sometimes predict one spectral line where there are actually two closely spaced spectral lines.

• This led to the proposal of a fourth quantum number, the spin quantum number m s .

• fflj can have two possible values for an electron. It can be 'spin-up' with m s = +1/2 or 'spin-down'
with m s = -1/2

• When two electrons occupy the same orbital, they must have different spin quantum numbers.
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• An orbital containing two electrons will have no net spin. When this is the case, the two electrons
are called diamagnetic electrons.

• An orbital containing only one electron will have a total spin equal to the spin of the electron that
it contains. When this is the case, the electron is called a paramagnetic electron.

• Electron spin helps to determine the magnetic properties of an atom.

• If all electrons in an atom are diamagnetic, the entire atom has no net spin, and is termed a 'dia-
magnetic atom.' Diamagnetic atoms are slightly repelled from a magnetic field.

• If an atom contains even one paramagnetic electron, the entire atom has a net spin and is termed a
paramagnetic atom. Paramagnetic atoms are slightly attracted to a magnetic field.

Review Questions

1. The principal quantum number describes the size of an electron energy level, the azimuthal quantum
number describes the shape of an electron energy level, and the magnetic quantum number describes
the orientation of the electron energy level. If there was another quantum number, what do you
think it might describe about the electron?

2. There is, in fact, a fourth quantum number that we'll learn about in this lesson. The fourth quantum
number is called the spin quantum number. Now can you guess what the final quantum number
might describe?

3. Choose the correct statement.

(a) The spin quantum number for an electron can only have the values m s = +1 and m s = — 1

(b) The spin quantum number for an electron can only have the value m s =

(d) The spin quantum number for an electron can only have the values m s = +1/2 and m s = -1/2

(e) The spin quantum number does not apply to electrons

4. Choose the correct statement.

(a) When two electrons share an orbital, they always have the same spin quantum numbers

(b) When two electrons share an orbital, they always have opposite spin quantum numbers

(d) When two electrons share an orbital there is no way to predict whether or not they will have
the same spin quantum numbers

5. Fill in the blanks in the following statement using numbers.

When scientists used the Schrodinger equation with only quantum numbers, they found that the

Schrodinger equation was pretty good at predicting atomic spectra, except that there were occasionally

closely spaced lines of light where the Schrodinger equation predicted only . This led scientists to

suggest that a complete description of an electron, which required quantum numbers.

6. In many atomic spectra, there are two very closely spaced lines of light which can only be predicted
by including the spin quantum number into the Schrodinger equation. Decide whether the following
statements about these two lines are true or false.

(a) the two lines spread further apart when the atom is placed in a magnetic field

(b) the two lines move closer together when the atom is placed in a magnetic field

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(c) the two lines are the result of an experimental error. If scientists are careful, they find that
there is really just one line.

(d) the two lines actually result from the fact that there are two very closely spaced energy states

7. Goudsmit and Uhlenbeck proposed the existence of

(a) the principal quantum number

(b) the azimuthal quantum number

(d) the magnetic quantum number

8. Circle all of the quantum numbers that tell you about the region in space where you're most likely
to find the electron.

(a) the spin quantum number

(b) the magnetic quantum number

(d) the azimuthal quantum number

9. Select the correct statement from the list below. An electron with a spin quantum number of
m s = -1/2

(a) cannot share an orbital with an electron that has a spin quantum number of m s = +1/2

(b) prefers to share an orbital with an electron that has a spin quantum number of m s = -1/2

(d) cannot share an orbital with another electron

10. What is the total spin in an electron orbital if

(a) the orbital contains one 'spin-up' electron

(b) the orbital contains one 'spin-down' electron

(d) .

Further Reading / Supplemental Links

• http : //www . ethbib . ethz . ch/exhibit/pauli/elektronenspin_e . html

• http : //www . lorentz . leidenuniv . nl/history/spin/goudsmit . html

• http://en.wikipedia.org/wiki

Vocabulary

spin quantum number, m s The fourth quantum number that must be included in the wave function of
an electron in an atom in order to completely describe the electron.

spin-up The term applied to electrons with spin quantum number m s = +1/2.

spin down The term applied to electrons with spin quantum number m s = -1/2.

diamagnetic electrons Two electrons with opposite spins, paired together in an orbital.
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diamagnetic atom An atom with no net spin; an atom with only diamagnetic electrons.

paramagnetic electron An electron alone in an orbital.

paramagnetic atom An atom with a net spin; an atom with at least one paramagnetic electron.

7.2 Pauli Exclusion Principle

Lesson Objectives

• Explain the meaning of the Pauli Exclusion Principle.

• Determine whether or not two electrons can coexist in the same atom based on their quantum
numbers.

• State the maximum number of electrons that can be found in any orbital.

Introduction

When electrons are found inside an atom, they're restricted to specific areas, or regions within the atom
which can be described by orbitals. Let's see what this means in terms of quantum numbers.

No Two Electrons in an Atom Can Have the Same Four Quantum
Numbers

How do you know that two electrons are in the same orbital? In order to fully specify an orbital, you need
to know the principal quantum number, n, the azimuthal quantum number, C, and the magnetic quantum
number, m\. The values of first three quantum numbers for an electron determine exactly which orbital the
electron in. Clearly, then, in order to be in the same orbital, two electrons have to have exactly the same
values for n, {, and m/. Now when two electrons have exactly the same values for n, t, and mi, they share
the same region of space within the atom, and in the last lesson, you learned that that had important
consequences in terms of their spins. If you remember back to an earlier section, electrons in the same
orbital, sharing the same region of space, had to have different values of m s . If one electron had m s = +1/2,
then the other had to have m s = -1/2 and vice versa. Let's take a look at several examples.

Example 1:

An electron with n = 2, t = 1, mi = — 1 and m s = +1/2 is found in the same atom as a second electron
with n = 2, t = 1, and m; = — 1. What is the spin quantum number for the second electron?

First electron: n = 1,1 = l,m\ = — l,m s = +1/2

Second electron: n = 1,1 = l,m/= —l,m s =?

Since the first three quantum numbers are identical for these two electrons, we know that they are in the
same orbital. As a result, the spin quantum number for the second electron cannot be the same as the spin
quantum number for the first electron. This means that the spin quantum number for the second electron
must be m s = -1/2.

Example 2:

An electron with n = 5, I = 4, m/ = 3 and m s = -1/2 is found in the same atom as a second electron with
n = 5, t = 4, and m/ = 3. What is the spin quantum number for the second electron?

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First electron: n = 5,-C = 4, mi = 3,m s = -1/2

Second electron: n = 5, t = 4, m; = 3,m s =?

Notice that whenever the two electron's first three quantum numbers are the same, the fourth is different.
Let's take a look at a few more examples...

Example 3:

Can an electron with n = 1, t = 0, mi = and m s = +1/2 exist in the same atom as a second electron with
n = 2, I = 0, mi = and m, = +1/2?

First electron: n = 1,1 = 0,w/ = 0,m s = +1/2

Second electron: n = 2,1 = 0, mi = 0,m s = +1/2

Since these two electrons are in different orbitals, they occupy different regions of space within the atom.
As a result, their spin quantum numbers can be the same, and thus these two electrons can exist in the
same atom.

Example 4:

Can an electron with n = 3, I = 1, mi = — 1 and m s = -1/2 exist in the same atom as a second electron
with n = 3, t = 2, mi = — 1 and m s = -1/2?

First electron: n = 3,£ = l,mi = —l,m s = -1/2

Second electron: n = 3, € = 2, mi = -l,m s = -1/2

Example 5:

Can an electron with n = 1, £ = 0, mi = and m s = +1/2 exist in the same atom as a second electron with
n = 2, t = 1, mi = and m s = +1/2?

First electron: n = 1,1 = 0,ra/ = 0,m s — +1/2

Second electron: n = 2, 1 = \,m\ = 0,m s = +1/2

Notice that whenever the two electrons have different values of n, or different values of {, or different values
of mi, they can have the same spin quantum number m s , because they are not in the same orbital, and
thus they are not sharing the same region of space within the atom. Let's take a look at one final example

Example 6:

Can an electron with n = 1, t = 0, mi = and m s = +1/2 exist in the same atom as a second electron with
n = l ; l = 0, mi = and m, = +1/2?

First electron: n = 1,1 = 0,ra/ = 0,m s = +1/2

Second electron: n = 2, 1 = l,m; = 0,m s = +1/2

Since these two electrons are in the same orbital, they occupy the same region of space within the atom.

www.ckl2.org 234

As a result, their spin quantum numbers cannot be the same, and thus these two electrons cannot exist in
the same atom.

Hopefully after having looked at seven different examples, it should be obvious to you that electrons in the
same atom with the same spin must be in different orbitals, while electrons in the same orbital of the same
atom must have different spins. As a result, no two electrons in the same atom can have exactly the same
four quantum numbers. If two electrons have the same n, the same €, and the same mi, then they are in
the same orbital. If they also have the same m s , then they also have the same spin, and that is impossible.

The first scientist to realize that two electrons in the same atom couldn't have the same four quantum
numbers was a man name Wolfgang Pauli (Figure 7.2). In 1925, Pauli stated what has come to be known
as the Pauli Exclusion Principle. The Pauli Exclusion Principle states that no two identical fermions
(a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum
state in an atom simultaneously. In other words, no two electrons in the same atom can have the same
four quantum numbers. If n, I, and mi are the same, m s must be different such that the electrons have
opposite spins.

* /is

■%
t

Figure 7.2: Wolfgang Pauli, the scientist who first proposed the Pauli Exclusion Principle.

No Atomic Orbital Can Contain More than Two Electrons

An electron can share its territory, or its orbital, with another electron, but only if the other electron is
slightly different - in other words, only if the other electron has a different spin.

There's a limit to the number of different electrons that can share an orbital, because there's a limit to the
number of different spins that those electrons can have. When it comes to spins, though, there are only
two possibilities. An electron can either be 'spin-up', with m s = +1/2, or 'spin-down', with m s = -1/2.
Therefore, if an orbital has one electron that is 'spin- up', and a second electron that is 'spin-down', the
orbital is full. What if a third electron tried to enter the orbital? Well, if the third electron was 'spin-up'
it would have trouble sharing the orbital, with the 'spin- up' electron that's already there. Similarly, if
the third electron was 'spin-down', it would have trouble sharing the orbital with the 'spin-down' electron
that's already there. Since the only two options for the third electron are 'spin-up' and 'spin-down', there's
really nothing that third electron can do - it just has to move on and find a new orbital! To summarize,
then, because there are only two possibilities for the spin quantum number of an electron,

NO ATOMIC ORBITAL CAN CONTAIN MORE THAN TWO ELECTRONS!

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Lesson Summary

• The Pauli Exclusion Principle states that 'no two identical fermions may occupy the same quantum
state in an atom simultaneously'. That is, no two electrons in an atom can have n, t, mi and m s all
the same.

• No atomic orbital can contain more than two electrons.

Review Questions

1. Electrons in the same orbital must have different spin quantum numbers. What is true of the other
three quantum numbers for two electrons in the same orbital?

2. Electrons in different orbitals can have the same spin quantum numbers. What is true of the other
three quantum numbers for two electrons in different orbitals?

3. Fill in the blank using either the word 'can', or 'cannot'.

(a) An electron with the quantum numbers n = 1, I = 0, m/ = and m s = +1/2 exist in

the same atom as an electron with the quantum numbers n = 2, £ = 0, m; = and m s = +1/2

(b) An electron with the quantum numbers n = 1, € = 0, mi = and m s = +1/2 exist in

the same atom as an electron with the quantum numbers n = 1, I = 0, mi = and m s = -1/2

4. Fill in the blanks using numbers.

(a) There is only 1 orbital at the n = 1 energy level. Therefore the n = 1 energy level can hold a
maximum of electrons

(b) There are 4 orbitals at the n = 2 energy level. Therefore the n = 2 energy level can hold a
maximum of electrons

(d) There are 16 orbitals at the n = 4 energy level. Therefore the n = 4 energy level can hold a
maximum of electrons

5. What is the maximum number of electrons that can exist in p orbitals at energy levels with n < 3.

6. What is the maximum number of electrons that can exist in p orbitals at energy levels with n < 5.

Further Reading / Supplemental Links

• http : //theory . uwinnipeg . ca/mod_tech/nodel68 . html

• http://en.wikipedia.org/wiki

Vocabulary

Pauli Exclusion Principle No two fermions may occupy the same quantum state in an atom simulta-
neously; no two electrons in an atom can have the same four quantum numbers.

7.3 Aufbau Principle

Lesson Objectives

• Explain the Aufbau Principle.

• Given two different orbitals, predict which the electron will choose to go into.

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Introduction

Everything in the universe is driven to minimize its potential energy. Previously we learned that heavy
objects fall when they're dropped, because their total potential energy is lower on the ground than it is
when they're hovering in the air. That's why Wile E. Coyote could drop an anvil on Road Runner -
he knew that the anvil would fall, because falling would lower its potential energy. It's the same with a
bowling ball placed on a hill. At the top of the hill, the bowling ball has more potential energy than it
does at the bottom, so you can always bet on the bowling ball rolling down the hill rather than up the hill,
because the bowling ball will always try to minimize its energy. In this chapter, we won't look at heavy
objects like bowling balls and anvils. Instead, we'll look at tiny objects - we'll look at electrons. Even
though electrons are much, much smaller than anvils and bowling balls, the same principle applies. An
electron will do anything that it can to lower its potential energy.

Electrons are Found in Energy Levels with Increasingly Higher
Energy

Whenever an electron is found inside an atom, it exists in what's known as an orbital. By now, you
should know what an orbital is. An orbital describes a particular region of space within the atom where
the electron is most likely to be found. While orbitals are important when it comes to figuring out an
electron's probable location, they are equally important when it comes to figuring out an electron's energy.
Electrons in different orbitals frequently have different energies. Of course, an electron is never going to
"choose" to be in an orbital that has a higher energy if there's space available in an orbital that has a
lower energy. It's a lot like you riding the bus. You would never choose to waste energy walking 20 miles
to school provided there was space for you on the bus. If there wasn't any space on the bus, though, you
may be forced to walk. It's the same with electrons.

As pictured below, if there's no room on the bus, you may be forced to walk. It's the same with electrons.
If there's no room in a low energy orbital, they may be forced into a higher energy orbital. (Source:
http : //en . wikipedia . org/wiki/Image : HDaumierOmnibus . JPG . Public Domain)

If there isn't any space in a low energy orbital, an electron may be forced into a higher energy orbital. The
fact that electrons always fill up lower energy orbitals first has important consequences when it comes to
determining which orbitals contain electrons in any given atom.

237

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Remember that the principal quantum number, n, is associated with the 'energy level' of the electron.
Electrons with standing waves described by bigger values of n had higher energies, while electrons with
standing waves described by smaller values of n had lower energies. If you were an electron, then, and
you had the choice of being in an orbital with n = 1 or an orbital with n = 2, which would you choose?
Obviously you'd choose to be in the orbital with n = 1, because it has a smaller value of n, and thus a
lower energy (remember, both people and electrons prefer to be in states with lower energy).

Naturally, there is a limit to the total number of electrons that can exist in the same atom at the n = 1
energy level. In fact, it turns out that there can be at most two electrons with n = 1 in any given atom.
That's because there is only one n = 1 orbital per atom. Of course, there are many atoms with more than
two electrons. Lithium, for instance, has three. What happens to the electrons in an atom like lithium?
Obviously, the first two electrons are going to occupy the single orbital that exists at the n = 1 energy
level. Since this orbital only has room for two electrons, though, the third electron has to move up to the
n = 2 energy level. In other words, electrons will fill up orbitals in order of increasing energy. If there's
space at the n = 1 energy level, that space will be filled before any electrons move into the n = 2 energy
level. Similarly, if there's space at the n = 2 energy level, that space will be filled before any electrons
move into the n = 3 energy level.

In an Energy Level Electrons are Assigned to Sublevels with In-
creasingly Higher Energy

So far you know that an electron will always be found occupying the orbital in the first energy level, rather
than the second energy level, provided that there is space available. Similarly, an electron will be found
occupying the second energy level, rather than the third energy level, provided that there is space available.
But what about electron sublevels? Among the orbitals with the same value of n, the s orbital will always
be filled first, followed by the p orbitals (and then the d orbitals).

We can summarize these rules with the following statement: Electrons will fill available orbitals starting
with those at the lowest energies before moving to those at higher energies. This statement is known as
the Aufbau Principle. That may sound like a funny name, but 'aufbau' is actually the German word
for 'construction', and the Aufbau Principle describes how the orbitals are 'constructed' by progressively
adding electrons to higher and higher energy levels.

Before we move on and consider exactly how many electrons go into each energy level, and sublevel, it's
important to point out how the energies of the electrons in an atom relate to the energy of the entire atom
itself. If each electron tries to minimize its energy by going into the lowest energy orbital available, then
the total energy of all the electrons in the atom is also as low as possible.

Let's compare electrons minimizing their energy in an atom to your relatives minimizing their energy
consumption in your family. If you do your best to turn off the lights and save as much energy as possible,
and your brother does his best to turn off the lights and save as much energy as possible, and your sister
does her best to turn off the lights and save as much energy as possible, and your parents also do their best
to turn off the lights and save as much energy as possible, then collectively, your entire family is saving as
much energy as possible as well. It's the same with electrons. If the first electron fills the lowest energy
orbital available to it, and the second electron fills the lowest energy orbital available to it, and the third
electron fills the lowest energy orbital available to it, and the fourth electron also fills the lowest energy
orbital available to it, then collectively, the entire atom is in the lowest energy state possible as well.

Just as you would prefer to minimize your energy by sleeping, or riding a bus, electrons minimize their
energy by occupying the lowest energy orbital available to them and atoms minimize their energy by having
all of their electrons in the lowest energy 'configuration' (arrangement) possible. Later, you will learn that
practically all chemical processes rely on this same principle of energy minimization.

www.ckl2.org 238

Lesson Summary

• In an atom, electrons will fill up orbitals in order of increasing energy.

• The principle quantum number determines the 'energy level' of the orbital. Orbitals with lower values
of n are usually associated with lower energy and will be filled first.

• The azimuthal quantum number determines the 'sublevel' of the orbital.

• Orbitals with lower values of t (but the same value of n) are always associated with lower energy and
will be filled first.

• The Aufbau Principle states that electrons will fill available orbitals starting with those at the lowest
energies before moving to those at higher energies.

• Since each electron in an atom minimizes its energy, the energy of the entire atom is a minimum as
well.

Review Questions

1. While we have talked about emission spectra, another type of spectra is known as absorption spectra.
In emission spectra, the atom emits lines of light like those you saw in the examples of atomic spectra.
In absorption spectra, the atom absorbs lines of light, rather than emitting them. Can you explain
this in terms of electrons and orbitals? What do you think the relationship between absorption
spectra and emission spectra might be?

2. If an electron has a "choice" between going into an orbital in the n = 1 energy level or an orbital in
the n = 2 energy level, which do you think it chooses?

3. If an electron in the n = 3 energy level has a "choice" between going into an orbital with t = or an
orbital with t = 1, which do you think it chooses?

4. Select the correct statement. According to the Aufbau Principle

(a) orbitals with higher values of n fill up first

(b) orbitals in the same energy level, but with higher values of t fill up first

(d) it is impossible to predict which orbitals will fill up first

5. Decide whether each of the following statements is true or false.

(a) Electrons in different orbitals have different energies.

(b) An electron will enter an orbital of higher energy when a lower energy orbital is already filled.

6. Does the electron in the hydrogen atom absorb or emit energy when it makes a transition between
the following energy levels:

(a) n = 2 to n = 4

(b) n = 6 to n = 5

7. Fill in the blanks. There is one s orbital, three p orbitals, and five d orbitals in the n = 3 energy level
of an atom. If a particular atom has a total of 5 electrons in the n = 3 energy level, then there are...

(a) electrons in the s orbital

(b) electrons in p orbitals

8. Fill in the blanks. There is one s orbital, three p orbitals, five d orbitals and 7 / orbitals in the n = 4
energy level of an atom. If a particular atom has a total of 7 electrons in the n = 4 energy level, then
there are...

(a) electrons in the s orbital

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(b) electrons in p orbitals

(d) electrons in / orbitals

9. According to the Aufbau rule, which of the following atoms has a sub-shell that is exactly half-filled?

(a) Ba

(b) Al

(d) As

(e) O

Further Reading / Supplemental Links

• http://www. iun.edu/~cpanhd/C101webnotes/modern-atomic-theory/aufbau-principle.html

• http : //www . avogadro . co . uk/light/auf bau/auf bau . htm

• http://en.wikipedia.org/wiki

Vocabulary

Aufbau principle Electrons will fill available orbitals starting with those at the lowest energy before
moving to those at higher energies.

7.4 Writing Electron Configurations

Lesson Objectives

• Figure out how many electrons can exist at any given sublevel.

• Figure out how many different sublevels can exist at any given energy level.

• Be able to write electron configuration of any element given the total number of electrons in that
element.

• Be able to write either orbital representations or electron configuration codes.

Introduction

It's pretty easy to say 'electrons will fill the lowest available energy orbital'. Figuring out what that lowest
available energy orbital is, though, can be quite a challenge. In this lesson, we're going to try to decide
exactly which orbitals get filled, and when. Most people who are just beginning to learn about quantum
chemistry tend to find orbital filling problems very confusing, so we'll take it slowly, and show you how
diagrams and rules can help you to remember the ways in which different orbitals are organized and filled

Electron Configurations

How do electrons fill the different energy levels, energy sublevels, and orbitals? To understand that question,
it helps a lot to look at a diagram. Figure 1 shows the first three energy levels (marked by the large
differently colored blocks), and the sublevels (separated by dotted lines) that are present in any atom. In

www.ckl2.org 240

Figure 1, each of the circles represents an orbital and, of course, each can hold a total of two electrons.
Notice the red n = 1 block contains only an s orbital. That orbital will hold the first two electrons in the
atom. Once the red n = \ block is entirely filled, electrons will start filling the orange n = 2 block. In the
orange n = 2 block, there are four different orbitals. The first orbital is an s orbital and the other three
are p orbitals. Once all four orbitals in the orange n = 2 block have been filled, electrons will start filling
the yellow n = 3 block. In the yellow n = 3 block, there are nine different orbitals. The first orbital is an
s orbital, the next three are p orbitals, and the last five are d orbitals. When filling the n = 3 block, the s
sublevel will always be filled first, since it is lowest in energy. Again, it can hold at most 2 electrons. After
that, the next six will fill the three p orbitals.

OOO

OOOOO

Figure 7.3: These are the first three energy levels with their sub-levels.

So far we've talked about filling the orbitals in Figure 7.3 up to the 3p orbitals. That's a total of 18
electrons. The first 18 electrons are 'nice and easy', because they fill the orbitals in order. First, all of the
n = 1 orbitals get filled, then all of the n = 2 orbitals get filled, then the n = 3 orbitals get filled. From
the 19 rfi electron on, though, things get a little crazy! Before we move on to the that region beyond 18
electrons, let's take a brief look at a shorthand notation that scientists use to signify the electron orbital
filling of a given atom.

Since orbitals are filled in order of increasing n and, within each energy level, in order of increasing £,
scientists can use a short hand, known as the electron configuration code, to represent filled orbitals. To
write the electron configuration code for an atom, you write the symbol for the type of orbital present at a
particular sublevel (Is, 2s, 2p, etc.) followed by a superscript to indicate how many electrons are actually
in that sublevel in the atom you are describing. Let's take a look at a few examples.

Example 1:

Nitrogen has 7 electrons. Write the electron configuration for nitrogen.

Take a close look at Figure 1, and use it to figure out how many electrons go into each sublevel, and also
the order in which the different sublevels get filled.

1. Begin by filling up the Is sublevel. This gives Is 2 . Now all of the orbitals in the red n = 1 block

are filled.
Since we used 2 electrons, there are 7-2 = 5 electrons left

2. Next, fill the 2s sublevel. This gives ls 2 2s 2 . Now all of the orbitals in the s sublevel of the orange

n = 2 block are filled.
Since we used another 2 electrons, there are 5-2 = 3 electrons left

3. Notice that we haven't filled the entire n = 2 block yet... there are still the p orbitals! The final 3

electrons go into the 2p sublevel. This gives ls 2 2s 2 2p^

241 www.ckl2.org

The overall electron configuration is: ls 2 2s 2 2p^ .

Overlapping Energy Levels

If you had to guess, which orbital do you think would be filled after the 3p orbitals? Most likely you'd
guess that the 3d orbitals would come next - and that guess makes a lot of sense. Unfortunately, that
guess is also wrong! It turns out that the 4s orbitals are filled before the 3d orbitals, even though the
4s orbitals have n = 4 and the 3d orbitals only have n = 3.

How is that possible? Does it mean that electrons go into higher energy orbitals before completely filling
the lower energy orbitals? Is there something wrong with those electrons? Are certain electrons prevented
from entering low energy orbitals? What's going on?

It turns out that there's nothing wrong with those 4s electrons at all. They're still behaving like normal
electrons and they're still going into the lowest energy orbital available. The only difference is that the 4s
orbital is a lower in energy than the 3d orbitals. Sometimes, we get tricked into thinking that the
principal quantum number determines which orbitals will get filled first. When it comes to the order in
which orbitals are filled, though, the principal quantum number isn't the only factor - what also matters
is the energy of the orbital. Usually orbitals with lower principal quantum numbers have lower energies,
but that isn't the case when you compare the energies of the 3d orbitals and the energies of the 4s orbitals.
In this case, the 4s orbitals are lower in energy than the 3d orbitals even though they have higher principal
quantum numbers. Figure 7.4 shows a modified version of Figure 7.3, which shows the n = 4 orbital
positions. Let's take a look at an example of an electron where some n = 4 orbitals are filled.

» OOOOO- -* d
O 5s

OOO 4 p

m OOOOO y

(J 4s

o ooo ;

o QQQ 2

-Is

Increasing
Energy

Figure 7.4: Notice that the orbital from the next higher energy level has slightly lower energy than the
orbitals in the lower energy level.

Example 2:

Potassium has 19 electrons. Write the electron configuration code for potassium.
This time, take a close look at Figure 7.4.

1. Begin by filling up the Is sublevel. This gives Is 2 . Now the n = 1 level is filled. Since we used 2

electrons, there are 19 - 2 = 17 electrons left

2. Next, fill the 2s sublevel. This gives ls 2 2s 2

Since we used another 2 electrons, there are 17 - 2 = 15 electrons left

3. Next, fill the 2p sublevel. This gives ls 2 2s 2 2p e . Now the n = 2 level is filled. Since we used another

6 electrons, there are 15-6 = 9 electrons left

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4. Next, fill the 3s sublevel. This gives ls 2 2s 2 2p 6 3s 2

Since we used another 2 electrons, there are 9-2 = 7 electrons left

5. Next, fill the 3p sublevel. This gives ls 2 2s 2 2p 6 3s 2 3p 6

Since we used another 6 electrons, there are 7-6=1 electron left

Here's where we have to be careful — right after 3/? 6 !!
Remember, 4s comes before 3d\

6. The final electron goes into the As sublevel. This gives ls 2 2s 2 2p e 3s 2 3p e As 1
The overall electron configuration code is: ls 2 2s 2 2p e 3s 2 3p e 4:S 1 .

The Diagonal Rule

Unfortunately, 4s orbitals aren't the only ones that get filled earlier than you'd expect based on their
principal quantum number. The same turns out to be true of the 5s orbitals as well. Even though 5s
orbitals have a higher principal quantum number than Ad orbitals, (n = 5 compared to n = 4), they're
actually lower in energy. As a result, 5s orbitals are always filled before 4d orbitals. Similarly, 6s orbitals
are lower in energy than 5d orbitals, so 6s orbitals are always filled first. The story gets even stranger
when you consider / orbitals. 5s, 5p and 6s orbitals are all lower than 4/ orbitals. In other words, before
you can get an electron into a 4/ orbital, you must first fill up the 5s orbitals, and the 5p orbitals and the
6s orbitals!

Filling up orbitals and writing electron configurations was so easy for atoms with less than 18 electrons!
But for atoms with more than 18 electrons, it seems hopeless to memorize all of the different rules. How
will you ever get straight whether the 5s orbital is higher or lower than the 4d orbital or the 4/ orbital?
Thankfully, there's a simple rule known as the diagonal rule. The diagonal rule states that: Electrons fill
orbitals in order of increasing 'quantum number sum' (n + €). When two orbitals share the same quantum
number sum, they will be filled in order of increasing n.

Luckily, the diagonal rule also has a diagram that's easy to remember, and that allows you to easily figure
out the order in which electron orbitals are filled. Figure ?? shows the diagram. In order to use it, you
must follow the arrows from tail-to-tip, starting with the first arrow in the upper left-hand corner, and
working your way down through the arrows to the lower right-hand corner of the diagram. To see how this
works, let's take a look at an example.

Example 3:

243

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Hafnium has 72 electrons. Write the electron configuration for hafnium.

We can follow the arrows in the diagram (as shown below), until we have assigned all 72 electrons. Notice
how you fill orbitals as you progress along the arrows, starting from the top arrow and moving down.
Remember to stop once you hit 72. At that point, you have finished writing the electron configuration for
Hafnium.

(total: 2 electrons) 1s
(total: 4 electrons) 1s 2s 2

(total: 12 electrons) 1s 2 2s2p3s 2

(total: 20 electrons) 1s 2 2s2p3s3p4s

(total: 38 electrons) Is^p^s^pVSdVss 2

(total: 56 electrons) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4cl5p 6 6s 2

(total: 72 electrons) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d ,0 4p 6 5s4d5p6s4l ?4 5d 2

Now that we know how electrons are assigned to orbitals, we can start to talk about how different orbitals,
and different electron configurations actually affect the chemical properties of different atoms. In other
words, we can actually start to talk about chemistry! Sometimes it seems that quantum physics and
quantum chemistry are very far removed from the real world chemistry that we see and use in every
day life. How are orbitals important when it comes to developing new drugs to treat cancer? How are
energy levels important when it comes to inventing different kinds of superconductors or plastics? How are
electron standing waves important when it comes to testing for toxins in your food? All of these processes
are determined by chemical properties which are, themselves, a direct result of how electrons are arranged,
and interact within different atoms and molecules. Sometimes it is easy to forget about the electrons, and
some of the strange quantum properties of subatomic particles. Nevertheless, we will never be able to
fully understand chemistry, if we don't understand its smallest components - components like electrons,
protons, neutrons and, of course, the atom.

Lesson Summary

• For any atom with less than 18 electrons, orbitals are filled in order of increasing n and, for any given
n, in order of increasing I.

• Electron configurations are a shorthand notation for representing the filled orbitals in a given atom.
They are written using the principal quantum number, n, for the energy level, the letter (s,p,d or /)
for the sublevel, and a superscript for the number of electrons in that sublevel.

• For atoms with more than 18 electrons, the orbitals are filled in order of increasing n (and in order
of increasing I for a given n) up to the 18 electron; however after the 18 f ' ! electron the 4s orbital are
filled before the 3d orbitals. This is because the 4s orbital has lower energy than the 3d orbital.

• The diagonal rule states that electrons fill orbitals in order of increasing 'quantum number sum'
(n + €). When two orbitals share the same 'quantum number sum,' they will be filled in order of
increasing n.

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244

Review Questions

1. Write the electron configuration for beryllium. Beryllium has 4 electrons.

2. Write the electron configuration for silicon. Silicon has 4 electrons.

3. Write the electron configuration for nitrogen. Nitrogen has 7 electrons.

4. Write the electron configuration for chromium. Chromium has 24 electrons.

5. Write the electron configuration for silver. Silver has 47 electrons.

Further Reading / Supplemental Links

Website with lessons, worksheets, and quizzes on various high school chemistry topics.

• Lesson 3-1 is on the Development of the Atomic Theory.

• Lesson 3-2 is on the Development of the Atomic Model.

• Lesson 3-3 is on Atomic Structure.

• Lesson 3-4 is on the Periodic Table.

• Lesson 3.6 is on Electron Configuration, http : //www . f ordhamprep . org/gcurran/sho/sho/lessons/
lesson31 .htm

Vocabulary

electron configuration A short hand notation to indicate the electron orbitals which are filled in a
particular atom.

diagonal rule The electrons fill orbitals in order of increasing 'quantum number sum' {n + l). When two
orbitals share the same 'quantum number sum', they will be filled in order of increasing n.

quantum number sum The sum of the principal quantum number, n, and the azimuthal quantum
number, t, for an electron. That is n + I.

Image Sources

(1) http: //en. wikipedia.org/wiki/File: Wolf gang_Pauli_young.jpg. Public Domain.

(2) http://en.wikipedia.0rg/wiki/Image:UhlenbeckKramersG0udsmit.jpg. Public Domain.

(3) Richard Parsons. . CC-BY-SA.

(4) Richard Parsons. . CC-BY-SA.

245 www.cki2.0rg

Chapter 8

Electron Configurations and
the Periodic Table

8.1 Electron Configurations of Main Group Ele-
ments

Lesson Objectives

• Explain how the elements in the Periodic Table are organized into rows and columns.

• Explain how the electron configurations within a column are similar to each other.

Introduction

It probably seems like all we've been spending a lot of time learning about protons... and neutrons... and
electrons... and electrons... and more electrons... so you might be wondering - when do we actually get to
study chemistry? When do we get to study reactions? When do we get to study explosions? When do
we get to study plastics, and medicines that can be made by combining different kinds of chemicals? The
answer is now. We're finally ready to discuss the chemical properties of the simplest chemicals out there
- we're finally ready to discuss the elements. Remember, you have learned that there were 117 different
kinds of atoms, and that each was known as an element. And you have learned that atoms of different
elements have different numbers of protons. Hydrogen has 1 proton (and 1 electron if it's neutral), helium
has 2 protons (and 2 electrons, if it's neutral), and lithium has 3 protons (and 3 electrons, if it's neutral).
Finally, you have seen examples of the Periodic Table. Scientists use the Periodic Table to summarize
information about all of the known elements that exist in our world.

In this lesson, you will learn why the Periodic Table (as shown below) has such an unusual shape.

(Source: CK-12 Foundation. CC-BY-SA)

www.ckl2.org 246

14 15 16

HYDROGEN

4A 5A 6A

HELIUM

LITHIUM

BERYLLIUM

8 9 10

BOHON

CARBON

NITROGEN

OXYGEN

FLUORINE

NEON

MAGNESIUM

SODIUM

1 SB 1

ALUMINUM

SILICON

PHOSPHORUS

SULFUR

CHLORINE

ARGON

POTASSIUM

CALCIUM

SCANDIUM

TITANIUM

VANADIUM

CHROMIUM

MANGANESE

IRON

COBALT

NICKEL

COPPER

ZINC

GALLIUM

GERMANIUM

ARSENIC

SELENIUM

BROMIUM

KRYPTON

RUBIDIUM

Strontium

YTTRIUM

ZIRCONIUM

NIORIUM

MPLYRDENUM

TECHNETIUM

RUTHENIUM

RHDDIUM

PALLADIUM

SILVER

CADMIUM

INDIUM

TIN

ANTIMONY

TELLURIUM

IODINE

XENON

57-71

La-Lu

CESIUM

BARIUM

LANTHANIOES

HAFNIUM

TANTALUM

TUNGSTEN

RHENIUM

OSMIUM

IRIDIUM

FLATIUM

GOLD

MERCURY

THALUUM

LEAD

BISMUTH

POLONIUM

ASTATINE

RADDN

89-103

Ac-Lr

104

105

106

107

108

109

110

111

112

113

Uut

114

Uuq

115

Uup

116

Uuh

117

Uus

118

Uuo

FRANCIUM

RADIUM

RUTHERFORDIUM

DUBNIUM

SEABORGIUM

BDHRIUM

HASSIUM

MEITNERIUM

,_,«,.•„,.

cor-ER-iciu...

UNUNTHIUM

UNUNQUADIUM

UNUNPENTIUM

_«!_

.»u„™.

UHUHOCTIUM

LANTHANIDES

LANTHANUM

CERIUM

MOTI-M

NEODYMIUM

PROMETHIUM

SAMARIUM

EUROPIUM

GADOLINIUM

TERBIUM

BVSPROSIUM

HDLMIUM

ERBIUM

THULIUM

YTTERBIUM

LUTETIUM

ACTINIDES

100

101

102

103

ACTINIUM

THORIUM

PROTACTINIUM

URANIUM

NEPTUNIUM

PLUTONIUM

AMERICIUM

CURIUM

BERKELIUM

CALIFORNIUM

EINSTEINIUM

.EPMIUM

MENOELEVIUM

N OBELI UM

•w

Now, what's the first thing you thought when you saw the Periodic Table? If you're like most people, the
first thing you thought was probably something like, 'Wow - that's a funny shape! Why is the Periodic
Table shaped like that? Why is it lower in the middle? Why is it higher on either end? Why is there
that odd- looking disconnected piece at the bottom? The Periodic Table doesn't look like a table at all!'
In this chapter, you'll begin to see why the Periodic Table has such a funny shape. It turns out that the
shape of the Periodic Table actually helps to tell us about the chemical properties of the different elements
that exist in our world. In this section, for example, you'll learn that elements in the same column of the
Periodic Table have similar chemical properties. Later we'll take a look at how elements in the same row
are related.

Group 1A Elements Have One s Electron

Remember that according to the Aufbau principle electrons are added to low energy orbitals first and then,
as the low energy orbitals are filled up, electrons go into higher and higher energy orbitals. When one
atom reacts with another atom in a chemical reaction, it's the high-energy electrons that are involved.

Since it's only the high-energy electrons that participate in a chemical reaction, it's only the high-energy
electrons that we will concern us when we want to determine the chemical properties of a particular element.
Just how 'high' in energy does an electron need to be to participate in a chemical reaction? Well, in most
chemical reactions, the only electrons involved are the electrons in the highest energy level. In other words,
the electrons with the highest value of n (the principal quantum number), participate in chemical reactions,
while the electrons with lower values of n are called "core electrons," are closer to the nucleus and, as a
result, don't get involved. The electrons with the highest value of n are known as valence electrons. Core
electrons are also referred as non-valence electrons. Two different elements have similar chemical
properties when they have the same number of valence electrons in their outermost energy level.

Elements in the same column of the Periodic Table have similar chemical properties. So what does that
mean about their valence electrons? You guessed it! Elements in the same column of the Periodic Table
have the same number of valence electrons - that's why they have similar chemical properties. Let's see if

247

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this is true for some of the elements in the first column of the Periodic Table.

Example 1:

Write the electron configuration for hydrogen (//).

First, you need to find hydrogen on the Periodic Table. Take a look at the Periodic Table above. You
know that hydrogen is in the first column, and if you look carefully, you'll see that hydrogen also happens
to be at the top of the first column. The Periodic Table tells you that the atomic number for hydrogen
is Z = 1, thus hydrogen has 1 proton. Neutral hydrogen will also have 1 electron. You need to write the
electron configuration for an atom with 1 electron.

As shown in the figure below, the diagonal rule applied to hydrogen (H).

(total: 1 electrons

principal quantum
number: n = 1

1 valence electron in an s orbital

Therefore, we write the electron configuration for H : Is 1 .

Remember, when you write electron configurations, the number out in front always indicates the principal
quantum number, n, of a particular orbital, thus Is 2 has n = 1, while 3s 1 has n = 3. What is the highest
principal quantum number that you see in hydrogen's electron configuration? It's n = 1, so all electrons
with n = 1 are valence electrons. Hydrogen has 1 valence electron in an s orbital.

Example 2:

Write the electron configuration for lithium (Li).

First, you find lithium on the Periodic Table. The Periodic Table tells you that the atomic number for
lithium is Z = 3, thus lithium has 3 protons. Neutral lithium will also have 3 electrons. You need to write
the electron configuration for an atom with 3 electrons.

As illustrated in the figure below, the diagonal rule applied to lithium (Li), non-valence electrons: Is 2 .

(total: 3 electrons

(total: 2 electrons) ls 2 -A

lis 2 2s 1 *2*^^*V^^'

principal quantum
number: n = 2

1 valence electron in an s orbital

Therefore, we write the electron configuration for Li : ls 2 2s 1 .

What is the highest principal quantum number that you see in lithium's electron configuration?

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248

It's n = 2, so all electrons with n = 2 are valence electrons, and all electrons with n < 2 are non-valence
electrons. Lithium has 1 valence electron in an s orbital.

Example 3:

Write the electron configuration for sodium (Na).

First, you find sodium on the Periodic Table. The Periodic Table tells you that the atomic number for
sodium is Z = 11, thus sodium has 11 protons. Neutral sodium will also have 11 electrons. You need to
write the electron configuration for an atom with 11 electrons.

As shown below, the diagonal rule applied to sodium (Na). non-valence electrons: ls 2 2s 2 2p e .

(total: 2 electrons) Is 2

(total: 4 electrons) Is 2 2s 2

(total: 11 electrons) Is 2 2s 2 2p 6 s 1

principal quantum
number: n = 3

1 valence electron in an s orbital

Therefore, we write the electron configuration for Na : ls 2 2s 2 2p e 3s 1 .

What is the highest principal quantum number that you see in sodium's electron configuration?

It's n = 3, so all electrons with n = 3 are valence electrons, and all electrons with n < 3 are non-valence
electrons. (Don't be fooled by the 2p 6 orbitals. Even though they are p orbitals, not s orbitals, they have
n = 2, so they are non-valence electrons!) Sodium has 1 valence electron in an s orbital.

If you look at the last line in Example 1, Example 2, Example 3 you should notice a pattern.

Hydrogen has 1 valence electron in an s orbital

Lithium has 1 valence electron in an s orbital

Sodium has 1 valence electron in an s orbital

In fact, all elements in the first column of the Periodic Table have 1 valence electron in an s orbital.

Therefore, we would expect all of these elements to have similar chemical properties - and they do.
(Hydrogen is special because it is the first element in the Periodic Table. As a result, hydrogen has only
one proton and one electron, which give it special chemical properties. Sometimes scientists don't include
hydrogen in the first column of the Periodic Table, but instead give it its own 'special' column to reflect
its special properties - we won't do that here, but you should realize that hydrogen does not have all the
same chemical properties as the rest of the elements in its column.)

The elements in the first column of the Periodic Table (other than hydrogen) are known as Group 1A
metals, or Alkali Metals. When you compare the chemical properties of these elements (lithium, sodium,
potassium, rubidium, cesium, and francium), what you'll notice is that they are all remarkably similar.
Group 1A elements are metals, silver-colored, and soft. These elements are extremely reactive. Several of
them explode if you put them in water.

As pictured below, notice how the elements lithium (Li), sodium (Na), and potassium (K) all look alike.
They are all soft, silver metals. Since Li,Na and K are all Group 1A metals, they all share similar chemical
properties. (Source: http://en.wikipedia.Org/wiki/File:Limetal.JPG, http://en.wikipedia.org/
wiki/File : Nametal . JPG . jpg, http : //en . wikipedia . org/wiki /Image : Kmetal . jpg . CC-BY-SA)

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And finally, because they are so reactive, Group 1A elements are not found in their elemental form in
nature - in other words, you don't find pure sodium or pure potassium in nature.

As shown below, the diagonal rule applied to rubidium (Rb). non- valence electrons: ls 2 2s 2 2p^3s 2 3p^As 2 3d w Ap^.

(total: 2 electrons) Is 2 —

Is 2 2s 2 -** rs -^-^^ ^^'

(total: 4 electrons)
(total: 12electrons)ls 2 2s 2 2
(total: 20electrons)ls 2 2s 2 2p 6 3s 2 3p 6 4s
(total: 37electrons)ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 s 1

principal quanturr
number: n = 5

1 valence electron in an s orbital

Group 2A Elements Have Two s Electrons

All of the elements in the first column of the periodic table have 1 valence electron in an s sublevel. How
do you think the elements in the second column of the periodic table differ? Let's find out by taking a
look at a few examples.

Example 4:

Write the electron configuration for beryllium (Be).

First, you find beryllium on the Periodic Table. The Periodic Table tells you that the atomic number for
beryllium is Z = 4, thus beryllium has 4 protons. Neutral beryllium will also have 4 electrons. You need
to write the electron configuration for an atom with 4 electrons.

Therefore, we write the electron configuration for Be : ls 2 2s 2 .

What is the highest principal quantum number that you see in beryllium's electron configuration?

It's n = 2, so all electrons with n = 2 are valence electrons, and all electrons with n < 2 are non-valence
electrons. Beryllium has 2 valence electrons in an s orbital.

As shown below, the diagonal rule applied to beryllium (Be), non- valence electrons: Is 2 .

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250

(total: 2 electrons) is 2
(total: 3 electrons) Is 2 2s 2

principal quantum
number: n = 2

2 valence electron in an s orbital

Example 5:

Write the electron configuration for magnesium (Mg).

First, you find magnesium on the Periodic Table. The Periodic Table tells you that the atomic number
for magnesium is Z = 12, thus magnesium has 12 protons. Neutral magnesium will also have 12 electrons.
You need to write the electron configuration for an atom with 12 electrons.

Therefore, the electron configuration for Mg : ls 2 2s 2 2p 6 3s 2 .

What is the highest principal quantum number that you see in strontium's electron configuration?

It's n = 3, so all electrons with n = 3 are valence electrons, and all electrons with n < 3 are non-valence
electrons. Magnesium has 2 valence electrons in an s orbital.

Notice that:

Beryllium has 2 valence electrons in an s orbital.

Magnesium has 2 valence electrons in an s orbital.

As pictured below, sports drinks frequently contain the electrolyte calcium chloride. (Source: CK-12
Foundation. CC-BY-SA)

You can probably guess the number and type of valence electrons in an atom of calcium (Co) , strontium
(Sr), barium (Ba) or radium (Ra). If you guessed 2 electrons in an s orbital, then you guessed right! All

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elements in the second column of the Periodic Table have 2 valence electrons in an s orbital.

The elements in the second column of the Periodic Table are known as Group 2A metals, or Alkaline
Earth Metals. As you might expect, because all Group 2A metals have 2 valence electrons in an s
orbital, they all share similar chemical properties. Group 2A elements are metals, silver colored and are
quite reactive though they are not nearly as reactive as the Group 1A elements.

Group 3A Elements Have s and lp Electrons

All of the elements in the first column of the Periodic Table have 1 valence electron in an s sublevel and all
of the elements in the second column of the Periodic Table have 2 valence electrons in an s sublevel. Can
you make any prediction about the valence electrons in the third column of the Periodic Table? Where
is the third column of the Periodic Table? It turns out that there are really two different 'third columns'
in the Periodic Table. Take a close look at the figure of the Period Table (the first figure of this lesson).
Can you spot the column labeled '3A'? Can you spot the column labeled '3B'? Notice that the smallest
atom in the '3B' column has Z = 21 (Scandium, Sc), while the smallest atom in the '3A' column has Z = 5
(Boron, B). [You need to note that there is an alternate way to name 3A elements; the can also be referred
to as group 13 since these elements are in the 13" 1 column of the Periodic Table.] Therefore, it obviously
makes sense to discuss the 3A column first. Let's figure out how many valence electrons atoms in the 3A
column have:

Example 6:

Write the electron configuration for boron (5).

The Periodic Table tells you that the atomic number for boron is Z = 5, thus boron has 5 protons. Neutral
boron will also have 5 electrons. You need to write the electron configuration for an atom with 5 electrons.

Therefore, the electron configuration for B : ls 2 2s 2 2p 1 .

What is the highest principal quantum number that you see in boron's electron configuration?

It's n = 2, so all electrons with n = 2 are valence electrons, and all electrons with n < 2 are non-valence
electrons. Both the electron in the 2p orbital and the electrons in the 2s orbital are valence electrons.
Boron has 2 valence electrons in an s orbital, and 1 valence electron in a p orbital, for a
total of 3 valence electrons.

As pictured below, the diagonal rule applied to boron (B).

(total: 2 electrons ) is 2

(total: 3 electrons) Is 2 2s 2

(total: 5 electrons) Is 2 2s 2 ^2p 1 '

principal quantum,
number: n = 2

3 valence electrons: 2 in an s orbital,
1 in a p orbital

Example 7:

Write the electron configuration for aluminum (A/).

The Periodic Table tells you that the atomic number for aluminum is Z = 13; thus neutral aluminum has
13 protons and 13 electrons. You need to write the electron configuration for an atom with 13 electrons.

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252

Therefore, the electron configuration for Al : ls 2 2s 2 2p (i 3s 2 3p 1 .

What is the highest principal quantum number that you see in aluminum's electron configuration?

It's n = 3, so all electrons with n = 3 are valence electrons, and all electrons with n < 3 are non-valence
electrons. Both the electron in the 2>p orbital and the electrons in the 3s orbital are valence electrons.
Aluminum has 2 valence electrons in an s orbital, and 1 valence electron in a p orbital, for
a total of 3 valence electrons.

From Example 7 and Example 8, we have:

Boron has 2 valence electrons in an s orbital and 1 valence electron in a p orbital

Aluminum has 2 valence electrons in an s orbital and 1 valence electron in a p orbital

In fact, all elements in the 3A column of the Periodic Table have 2 valence electrons in an s orbital
and 1 valence electron in a p orbital. That's a total of 3 valence electrons for atoms in the 3A column.
Again, the chemical properties of 3A elements are similar, because they have the same number and type
of valence electrons.

Group 4A-8A Continue to Add p Electrons to the Outermost
Energy Level

By now, you may have noticed a pattern relating the number of valence electrons to the column number.
Group 1A elements have 1 valence electron. Group 2A elements have 2 valence electrons. Group 3A
elements have 3 valence electrons. Group 4A elements have... well, we haven't looked at them yet, but
what would you guess? It's pretty obvious. Group 4A elements have 4 valence electrons. Similarly, Group
5 A elements have 5 valence electrons. In fact, the pattern continues all the way up to Group 8 A elements,
which have 8 valence electrons. Let's take a look at a few examples in order to figure out exactly what
types of valence electrons are involved. First, we'll consider a Group 4A element.

Example 8:

Write the electron configuration for carbon (C).

The Periodic Table tells you that the atomic number for carbon is Z = 6, thus neutral carbon has 6 protons
and 6 electrons. You need to write the electron configuration for an atom with 6 electrons.

Therefore, the electron configuration for C : ls 2 2s 2 2p 2 .

What is the highest principal quantum number that you see in carbon's electron configuration?

It's n = 2, so all electrons with n = 2 are valence electrons, and all electrons with n < 2 are non-valence
electrons. Both the electrons in the 2p orbitals and the electrons in the 2s orbital are valence electrons.
Carbon has 2 valence electrons in an s orbital, and 2 valences electron in p orbitals, for a
total of 4 valence electrons.

Illustrated below, the diagonal rule applied to carbon (C). non-valence electrons: Is 2 .

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(total: 2 electrons) Is

(total: 4 electrons) Is 2 2s 2

(total: 12 electrons) Is 2 s 2 p

principal quantum
number: n = 2

4 valence electrons: 2 in an s orbital,
and 2 in p orbitals

Now let's consider a Group 5 A element.

Example 9:

Write the electron configuration for nitrogen (N).

The Periodic Table tells you that the atomic number for nitrogen is Z = 7, neutral nitrogen has 7 protons
and 7 electrons. You need to write the electron configuration for an atom with 7 electrons.

Therefore, the electron configuration for N : ls 2 2s 2 2p 3 .

What is the highest principal quantum number that you see in phosphorus's electron configuration? It's
n = 2, so all electrons with n = 2 are valence electrons, and all electrons with n < 2 are non- valence
electrons. Both the electrons in the 2p orbitals and the electrons in the 2s orbital are valence electrons.
Nitrogen has 2 valence electrons in an s orbital, and 3 valence electrons in p orbitals for a
total of 5 valence electrons.

As a final example, let's take a look at a Group 6A element (or Group 16).

Example 10:

Write the electron configuration for oxygen (0).

The Periodic Table tells you that the atomic number for oxygen is Z = 8; neutral oxygen has 8 protons
and 8 electrons. You need to write the electron configuration for an atom with 8 electrons.

Therefore, the electron configuration for O : ls 2 2s 2 2/A

What is the highest principal quantum number that you see in oxygen 's electron configuration?

It's n = 2, so all electrons with n = 2 are valence electrons, and all electrons with n < 2 are non-valence
electrons. Both the electrons in the 2p orbitals and the electrons in the 2s orbital are valence electrons.
Oxygen has 2 valence electrons in an s orbital, and 4 valence electrons in p orbitals, for a
total of 6 valence electrons.

So let's summarize what we know so far:

Group 1A elements have 1 valence electron in an s orbital

Group 2A elements have 2 valence electrons in an s orbital

Group 3A elements have 2 valence electrons in an s orbital and 1 valence electron in a p orbital

Group 4A elements have 2 valence electrons in an s orbital and 2 valence electrons in p orbitals

Group 5A elements have 2 valence electrons in an s orbital and 3 valence electrons in p orbitals

Group 6A elements have 2 valence electrons in an s orbital and 4 valence electrons in p orbitals

Can you guess how the valence electrons in columns 7A (Group 17) and 8A (Group 18) are arranged?

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Group 7A elements have 2 valence electrons in an s orbital and 5 valence electrons in p orbitals

Group 8A elements have 2 valence electrons in an s orbital and 6 valence electrons in p orbitals

Notice that, after column 3A, each column one step further to the right has one additional valence p
electron. Group 4A elements have one more valence p electron than Group 3A elements. Similarly, Group
5 A elements have one more valence p electron than Group 4A elements. But what happens when you
reach Group 8A elements? Why does the Periodic Table end at column 8A? Let's think about that
carefully. Group 8A elements have 6 valence electrons in p orbitals. In the last chapter, you learned that
the maximum number of p electrons at any energy level is 6. Therefore, there couldn't be a '9A' column,
because a '9A' column would have Ip electrons in the valence energy level, which is impossible.

The fact that Group 8A elements have completely filled valence s sublevel and p sublevel is important in
terms of their chemical properties. Group 8A elements are called Noble Gases.

They are all gases, and they are not very reactive at all.

Lesson Summary

• All known elements are organized into the Periodic Table in such a way that elements in the same
column have similar chemical properties.

• Only the highest energy electrons (valence electrons) are involved in chemical reactions. Therefore, it
is only these high-energy electrons that are important in determining an elements chemical properties.

• Two different elements are likely to have similar chemical properties when they have the same number
of valence electrons.

• Elements with the same number of valence electrons are found in the same column of the Periodic
Table.

• All elements in the first column of the Periodic Table have 1 valence electron in an s orbital. These
elements are known as Group 1A metals or alkali metals.

• All elements in the second column of the Periodic Table have 2 valence electrons in an s orbital.
These elements are known as Group 2A metals or alkaline earth metals.

• All elements in column 3A of the Periodic Table have 2 valence electrons in an s orbital and 1 valence
electron in a p orbital.

• All elements in column 4A of the Periodic Table have 2 valence electrons in an s orbital and 2 valence
electrons in p orbitals. ..etc.

• Column 8A has 2 valence electrons in an s orbital and 6 valence in p orbitals. Since any given energy
level can have at most dp electrons, column 8A elements have a filled p sublevel. Therefore, they are
inert (non-reactive), because they are unlikely to either gain or lose electrons. Group 8A elements
are called Noble Gases.

Review Questions

1. Take a look at the Periodic Table. How would you describe it? Why do you think it has such a funny
shape?

2. Can you suggest how elements in the same column of the Periodic Table might be similar?

3. Choose the correct statement.

(a) Mg has only 1 valence electron in an s orbital

(b) F has only 1 valence electron in an s orbital

(d) Kr has only 1 valence electron in an s orbital

(e) Fr has only 1 valence electron in an s orbital

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4. Circle the appropriate element for each blank.

(a) (Mg/N) has 2 valence electrons in an s orbital, and 3 valence electrons in

p orbitals.
(b) (As/B) has 2 valence electrons in an s orbital, and 3 valence electrons in

p orbitals
(c) (Cl/P/Li) has 2 valence electrons in an s orbital, and 5 valence electrons

in p orbitals
(d) {Al/Li/Na) has 1 valence electron in a p orbital

5. Choose the correct statement.

(a) Group 1A elements have a total of 3 valence electrons

(b) Group 5A elements have a total of 2 valence electrons

(d) Group 8A elements have a total of 8 valence electrons

(e) Group 2A elements have a total of 5 valence electrons

(f) Group 1A elements have a total of 3 valence electrons

6. Fill in the blanks.

(a) N has valence electrons in an s orbital

(b) ,/V has valence electrons in p orbitals

(d) Ca has valence electrons in s orbitals

(e) Ca has valence electrons in p orbitals

(f) Ca has a total of valence electrons

7. Decide whether each of the following statements is true or false.

(a) K has 1 valence electron in an s orbital

(b) Ge has 2 valence electrons in an s orbital

(d) B has 3 valence electrons in p orbitals

(e) F has 2 valence electrons in an s orbital, and 7 valence electrons in p orbitals

(f) Ca has a total of 4 valence electrons

8. Match the element to its valence electrons.

(a) Sr - i. a total of 8 valence electrons

(b) / - ii. a total of 2 valence electrons

(d) N - iv. a total of 7 valence electrons

9. Fill in the blanks.

(a) Ba has valence electron(s) in an s orbital, and valence electron(s) in p orbitals

(b) Sn has valence electron(s) in an s orbital, and valence electron(s) in p orbitals

(d) Po has valence electron(s) in an s orbital, and valence electron(s) in p orbitals

(e) Na has valence electron(s) in an s orbital, and valence electron(s) in p orbitals

10. List all of the elements with exactly 2 valence electrons in p orbitals.

11. An element has 2 valence electrons in an s orbital and 4 valence electrons in p orbitals. If the element
is in the second row of the Periodic Table, which element is it?

12. An element has 2 valence electrons in an s orbital and 6 valence electrons in p orbitals. If the element
is in the same row as In, which element is it?

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Further Reading / Supplemental Links

• http : //www . wou . edu/las/physci/ch412/perhist .htm

• http : //www . aip . org/history/curie/periodic . htm

• http: //web. buddyproject . org/web017/web017/history .html

• http://www.dayah.com/periodic

• http : //www . chemtutor . com/perich . htm

Vocabulary

Periodic Table Scientists use the Periodic Table to summarize what they know about the existing
elements. Elements of similar size are found in the same row, while elements with similar chemical
properties are found in the same column.

chemical properties The ways in which an element reacts with another element or compound.

valence electrons The electrons in an atom with the highest value of n (the electrons in the highest
energy level).

no n- valence electrons All electrons in atom which are not valence electrons. Non- valence electrons are
not important in determining an element's chemical properties because they rarely get involved in
chemical reactions.

alkali metals Group 1A metals. These are elements found in the first column of the Periodic Table,
excluding hydrogen.

alkaline earth metals Group 2A metals. These are elements found in the second column of the Periodic
Table.

noble gases Group 8A elements. These are elements found in the eight column of the Periodic Table.
They are inert, which means that they are very non-reactive.

8.2 Orbital Configurations

Lesson Objectives

• Draw orbital diagrams.

• Define Hund's Rule.

• Use Hund's Rule to decide how electrons fill sublevels with more than one orbital.

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Introduction

In the Electron Configurations of Main Group Elements lesson, you learned a little bit about valence
electrons. You saw how the number and type of valence electrons are important in determining the
chemical properties of a particular element. Group 1A metals were highly reactive, because they have a
strong tendency to lose their single valence s electrons. Group 2A metals are reactive as well, but less
so, because they had 2 valence s electrons. Finally, Group 8A elements were inert (not reactive at all),
because they had completely filled valences and p sub levels, meaning they could neither lose nor gain
electrons very easily. Now you might be wondering why we didn't talk much about the chemical properties
of the elements in columns 4A-7A. It turns out that understanding the behavior of these elements requires
a bit more information. Specifically, we need to know how the electrons fill up the p orbitals. Carbon,
for instance, is a Group 4A element, so it has 2 valence s electrons, and 2 valence p electrons. Obviously,
the 2 valence s electrons are paired together in the s orbital, but what about the 2 valence p electrons?
Are the valence p electrons paired in a single p orbital, or are they each in their own p orbital (remember,
there are a total of three p orbitals that the valence p electrons could be found in). What about nitrogen?
Nitrogen is a Group 5A element, so it has 2 valence s electrons, and 3 valence p electrons. Again, the 2
valence s electrons must be paired in the s orbital, but what about the 3 valence p electrons? Are two of
them paired in a single p orbital, or do all three have their own p orbitals?

Orbital Representation

Before we discuss the order and manner in which the orbitals in a p sublevel are filled, we have to introduce a
symbolic notation that scientists use to show orbital filling. Frequently, scientists will use boxes to represent
orbitals.

The figure below shows a set of boxes for orbitals up to the n = 2 energy level. The boxes are depicting
electron orbitals for the n = 1 and n = 2 energy levels. Notice that for n = 1, there is a single s orbital,
while for n = 2 there is one s orbital and three p orbitals.

Is 2s 2p

When a 'spin-up' electron is present in an orbital, scientists draw an upward pointing arrow in that orbital's
box. This leads to what is known as an orbital diagram. For example, hydrogen has a single electron in
the Is orbital. If that single electron were a spin-up, the orbital diagram for hydrogen would be:

Illustrated below, the orbital diagram for hydrogen, assuming that hydrogen's single electron is 'spin-up'
(m, = +1/2).

Is 2s 2p

When a 'spin-down' electron is present in an orbital, scientists draw a downward pointing arrow in that
orbital's box. Helium, for instance, has two electrons in the Is orbital. We know that one of these electrons
must be 'spin-up' and the other must be 'spin-down,' so the orbital diagram for helium would be:

As shown below, the orbital diagram for helium. Notice that when electrons are paired in an orbital, one
of the electrons is 'spin- up,' while the other is 'spin-down.'

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By comparing the first two orbital representation figures, it should be fairly obvious to you, which electrons
are paired and which are unpaired. That is one of the advantages of an orbital diagram. An orbital diagram
is a clear way of showing exactly how many paired and unpaired electrons there are in a particular atom's
electronic configuration.

Now you might be wondering about the unpaired electrons - how do you know whether to draw them
as 'spin-up,' or 'spin-down.' Technically speaking, the first electron in a sublevel (Is, 2s, 2p, etc) could be
either 'spin-up' or 'spin-down.' In other words, for hydrogen (Is 1 ), you could draw the arrow in the Is
orbital box pointing either up or down. Similarly, for boron (ls 2 2s 2 2p 1 ), you could draw the arrow in the
2p orbital box pointing either up or down. Once you've chosen the spin of the first electron in a sublevel,
though, the spins of all of the other electrons in that sublevel depend on the spin you chose for the first.
To avoid confusion, scientists 'always draw the first electron in an orbital as 'spin-up." If you stick with
this rule, you'll never get into trouble.

If you insist on breaking this rule, you might draw an orbital diagram that is incorrect, as shown in the
figure below. By convention, scientists usually draw all unpaired electrons as 'spin-up.' This prevents them
from drawing incorrect orbital diagrams like the one shown in (c). In the next section, you'll learn that
the orbital diagram in (c) is incorrect because it does not obey Hund's Rule.

I E

Is 2s

correct

"■IS

correct

(since this is the 1 electron in the p sublevel,
it can be either 'spin up' or 'spin down')

ffl

incorrect

(the spin of the 2 electron depends on the spin of the
f'electron, even though they aren't in the same orbital!

ffl

correct

(as long as you draw the//rst electron in each orbital as
'spin-up', you will draw a correct orbital diagram )

Notice that by drawing the first electron in each orbital as 'spin-up,' all of the unpaired electrons in the
sublevel have the same spin, even if they are in different orbitals! This is due to a principle known as
Hund's Rule. We'll discuss Hund's Rule in more detail in the next section.

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Orbitals in p Sublevel Fill Individually Before Pairing

Previously we learned that the energy of an electron in any given orbital depends on the energy level of the
orbital (which is determined by the principal quantum number, n) and the sublevel of the orbital (s,p,d,
etc. which is determined by the azimuthal quantum number: €). We also learned that according to the
Aufbau principle, electrons will fill the lowest energy orbitals first, and then move up to higher energy
orbitals only after the lower energy orbitals are full. If you think carefully, though, you'll realize that
there's still a problem. Certainly, Is orbitals should be filled before 2s orbitals, because the Is orbitals
have a lower value of n, and thus a lower energy. Similarly, 2s orbitals should be filled before 2p orbitals,
because 2s orbitals have a lower value of l{t = 0), and thus a lower energy. What about the three different
2p orbitals? In what order do electrons fill the 2p orbitals? To answer this question, we need to turn to a
principle known as Hund's Rule. Hund's Rule states that: (1) Every orbital in a sublevel is singly occupied
before any orbital is doubly occupied. (2) All of the electrons in singly occupied orbitals have the same
spin.

According to the first rule, electrons will always occupy an empty orbital before they pair up. This should
make sense given what you know about electrons. Electrons are negatively charged and, as a result, they
repel each other. Since electron-electron repulsion raises the energy of the electrons involved, electrons
tend to minimize repulsion (and thus minimize their energies) by occupying their own orbital, rather than
sharing an orbital with another electron. Take a look at the figure below.

no clcctron-clcetrcn repulsion
equals lower energy

11 si eh

electron- electron repulsion
equals higher energy

1*1

e a

2s
incorret

2p
t

Is 2s 2p

correct

Notice how the two 2p electrons in the orbital diagram on the left are in separate orbitals, while the two 2p
electrons in the orbital diagram on the right are sharing a single orbital. The orbital diagram on the left
is the correct orbital diagram, because it obeys Hund's Rule, meaning that there is less electron-electron
repulsion and, as a result, the electrons have lower energies (remember, electrons always minimize their
energies).

no elcetron-elcetroi repulsion
equals lower energy

m m Ltun

electron- electron repulsion
equals higher energy

1*1

E St

2s 2p
incorrect

Is 2s 2p

correct

The figure above illustrating orbital diagrams for nitrogen is similar to the orbital diagram for carbon in
the first figure. Notice how all three 2p electrons in the orbital diagram on the left are in separate orbitals,
while two of the three 2p electrons in the diagram on the right are sharing a single orbital. The orbital
diagram on the left is the correct orbital diagram, because it obeys Hund's Rule. Again, this means that
there is less electron-electron repulsion and, as a result, the electrons have lower energies.

no electron-electron repulsion
equals lower energy

ffl

electron- electron repulsion
equals higher energy

* BE i»i_t

E 5>

Is 2s 2p

2s 2p

correct

incorrect

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260

The figure above shows possible electron configurations for an atom with four 2p electrons. This time, two
of the electrons have no choice - they must pair up. The other two electrons, however, could either pair
up, as shown in the orbital diagram on the right, or occupy their own orbitals, as shown in the orbital
diagram on the left. Which do you think is correct? Obviously, the orbital diagram on the left, because it
minimizes electron-electron repulsion. The orbital diagram on the left is also the orbital diagram, which
follows Hund's Rule, since all orbitals are singly occupied before any are doubly occupied. The orbital
diagram on the right does not follow Hund's Rule, since the first two orbitals are doubly occupied before
the third is singly occupied.

While it's easy to understand why electrons would occupy empty orbitals before pairing up, it's a lot harder
to understand why unpaired electrons in different orbitals must all have the same spin. Electron spins in
different orbitals align (all point in the same direction), because spins, which are aligned have lower energy
than spins which are not aligned. Notice that as long as you always draw the first electron in an
orbital as 'spin-up' you will always draw spins which are aligned. Refer back to the figure illustrating
unpaired electrons as 'spin-up.' Now that you know Hund's Rule, it should be obvious why the orbital
diagram in 4.c is incorrect - the two electrons in the singly occupied 2p orbitals have different spins, and
thus this orbital diagram does not obey Hund's Rule. Compare the orbital diagram in 4.c to the orbital
diagram in 4. d. The orbital diagram in 4. d does obey Hund's Rule, because the two electrons in the singly
occupied 2p orbitals have the same spin.

Nitrogen Has Three Unpaired Electrons

Now that we know about orbital diagrams and Hund's rule, we can begin to explain the chemistry of the
elements in Groups 4A through 7A. Let's take a look at nitrogen as an example.

Example 1:

Draw the orbital diagram for nitrogen.

First, we need to write the electron configuration for nitrogen just as we did previously: which gives
ls 2 2s 2 2p' i . To draw the orbital diagram we will write the following: the first two electrons will pair up in
the Is orbital; the next two electrons will pair up in the 2s orbital. That leaves 3 electrons, which must
be placed in the 2p orbitals. According to Hund's Rule, all orbitals will be singly occupied before any
is doubly occupied. Therefore, we know that each p orbital gets one electron. Hund's Rule also tells us
that all of the unpaired electrons must have the same spin. Keeping with convention, we draw all of these
electrons as 'spin-up,' which gives.

The first two electrons pair The last three electrons singly occupy the three

up in the Is orbital 2porbitals. They all have the same spin!

\ J

1 s] rrrrrr

Is 2s 2p

The second two electrons
pair up in the 2s orbital

Orbital diagrams can help you to make predictions about the ways in which certain elements will react,
and the chemical 'compounds' or 'molecules' that different elements will form. We aren't ready to discuss
compounds, or molecules yet. The principles that you're learning now will help you to understand the
behavior of all chemicals, from the most basic elements like hydrogen and helium, to the most complex
proteins (huge biological chemicals made of thousands of different atoms bound together) found in your
body.

261 www.ckl2.org

Lesson Summary

• Orbital diagrams are drawn by representing each orbital as a box, each 'spin-up' electron in an orbital
as an upward pointing arrow in the box, and each 'spin-down' electron in an orbital as a downward
pointing arrow in the box. You can only have two arrows in each box, and they must be pointing in
opposite directions.

• Scientists use the convention that the first electron in any orbital is 'spin-up,' therefore, the first
arrow in an orbital 'box' should point up. Hund's Rule states:

(1) Every orbital in a sublevel is singly occupied before any orbital is doubly occupied.

(2) All of the electrons in singly occupied orbitals have the same spin.

• Electrons will occupy separate orbitals rather than pairing up, since this minimizes electron-electron
repulsions, thereby minimizing energy. Electron spins in different orbitals within the same sublevel
align because aligned spins have lower energy.

Review Questions

1. Which of the following is a valid orbital diagram?

T 1 T

1 T T

Tl TT T

tl T T

2. Draw the orbital diagram for lithium (Li).

3. Draw the orbital diagram for carbon (C).

4. Draw the orbital diagram for fluorine (F).

5. Draw the orbital diagram for oxygen, O. Use it to answer the following questions:

(a) an oxygen atom has

(b) an oxygen atom has

(d) an oxygen atom has

_ unpaired valence electrons
_ paired valence electrons
_ paired non-valence electrons
_ unpaired non-valence electrons

6. Draw the orbital diagram for neon, Ne. Use it to answer the following questions:

(a) a neon atom has unpaired valence electrons

(b) a neon atom has paired valence electrons

(d) a neon atom has unpaired non- valence electrons

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262

7. Decide whether each of the following statements is true or false.

(a) Every orbital in a sublevel is doubly occupied before any orbital is singly occupied.

(b) Every orbital in a sublevel is singly occupied before any orbital is doubly occupied.

(d) The two electrons in a single orbital have the same spin.

(e) All electrons in singly occupied orbitals have different spins.

(f) The two electrons in a single orbital have different spins.

8. Draw the orbital diagram for phosphorus P.

9. Draw an orbital diagram for silicon, Si. Use it to answer the following questions:

(a) a silicon atom has unpaired valence electrons

(b) a silicon atom has paired valence electrons

(d) a silicon atom has unpaired non-valence electrons

10. Draw an orbital diagram for Mn. Use it to determine the total number of unpaired electrons in an
Mn atom.

Further Reading / Supplemental Links

• http://www. chempractice. com/drills/ java_AO.php

• http : //library . thinkquest . org/3659/structures/electronconf ig . html

• http : //www . chem.latech.edu/~deddy/Lectnote/Chap7B . html

• http://www.mi .mun.ca/users/edurnf or/1100/atomic°/ 20structure/sld036.htm

Vocabulary

orbital diagram Orbital diagrams are drawn by representing each orbital as a box, each 'spin-up' elec-
tron in an orbital as an upward pointing arrow in the box, and each 'spin-down' electron in an orbital
as a downward pointing arrow in the box.

Hund's rule Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of
the electrons in singly occupied orbitals have the same spin.

8.3 The Periodic Table and Electron Configura-
tions

Lesson Objectives

• Relate an element's position in the PT to the energy level of its valence electrons (excluding transition
metals, lanthanides, and actinides).

• Relate an element's position in the PT to the sublevel of its highest energy valence electrons.

• Explain why there are only two elements in the first row of the PT.

263 www.ckl2.org

Introduction

With what we have already discussed, you might realize that just as electron configurations can be used
to explain the shape and organization of the Periodic Table, the shape and organization of the Periodic
Table can, in turn, be used to predict electron configurations. In fact, if you can locate an element on the
Periodic Table, you can use the element's position to figure out the energy level of the element's valence
electrons. Furthermore, an element's position on the Periodic Table tells you the sublevel of the element's
highest energy valence electrons. In this lesson, we'll take a close look at how the Periodic Table relates to
the electron configurations.

Rows Across on the PT are Consistent With the Energy Level in
An Atom

First, let's try to figure out what we can learn from an element's row or period in the Periodic Table.

The figure below shows how the different rows in the Periodic Table are numbered. The transition metals
and the lanthanides and actinides have been omitted. (Source: CK-12 Foundation. CC-BY-SA)

18
8A

[1.00784; 1.00811]

HYDROGEN

2 13 14 15 16 17
2A 3A 4A 5A 6A 7A

4.0026

HEUUM

; [6.938; 6.997]

LITHIUM

9.0122

BERYLLIUM

[10.S06; 10.821]

BORON

[12.0096; 12.0116]

CARBON

NITROGEN

[15.99903:15.99977]

OXYGEN

FLUORINE

20.180

NEON

22.990

SODIUM

MAGNESIUM

25.9B2
ALUMINUM

[28.084; 28.086]

SILICON

30.974

PHOSPHORUS

[32.059; 32.075]
SULFUR

[35.446; 35.457]

CHLORINE

39.94B

ARGON

39.09B

POTASSIUM

40.078

CALCIUM

69.723

GALLIUM

69.723
GERMANIUM

74.922
ARSENIC

7B.963

SELENIUM

79.904
BR0MIUM

83,801
KRYPTON

85.46S

RUBIDIUM

87.62

STRONTIUM

INDIUM

TIN

121.760

ANTIMONY

127.603

TELLURIUM

126.904

IODINE

131.292

XENON

132.905

CESIUM

137.327

BARIUM

[204.3B2; 204.385]

THALLIUM

204.383

LEAD

208.980

BISMUTH

20B.98?

POLONIUM

209.9B7

ASTATINE

222.018

RADON

223.020

FRANCIUM

226.0254

RADIUM

113

Uut

284

UNUNTRIUM

114

Uuq

284

UNUNQUAI1IUM

115

Uup

288
UNUNPENTIUM

116

Uuh

292

UNUNHFXIUM

117

Uus

294
UNUNSKPilUM

118

Uuo

294
UNUN0CTIUM

ROW1

ROW 2

ROW 3

ROW 4

ROWS

ROW 6

ROW 7

To understand what this means in terms of an element's electron configuration, let's consider the Group
1A metals. If we write the electron configuration for the Group 1A metal from each row of the Periodic
Table, we have:

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264

row
row
row
row
row
row

2
3

5
6

Li:

l5 2 2/

ls 2 2s 2 2p 6 3s l

ls 2 2s 2 2p 6 3s 2 3p 6 4s 1

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1

ls 2 2s 2 2p 6 3s 2 3 P Hs 2 3d l0 4p 6 5s 2 4d l0 5p%s l

s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d w 4p 6 5s 2 4d 10 5p%s 2 4f u 5d w 6p 6 7s 1

Do you see any pattern? For Group 1A metals, it seems that element's row corresponds to the energy level
of that element's valence electron. Lithium {Li), for instance, is found in row 2 of the Periodic Table, and
its valence electron is found in the n = 2 energy level. Cesium {Cs) is found in row 6 of the Periodic Table,
and its valence electron is found in the n = 6 energy level. Let's see if this same pattern holds for Group
2 A metals:

row
row
row
row
row
row

2
3

5
6

Li:

Na
K:
Rb
Cs
Fr

ls 2 2s 2

ls 2 2s 2 2p 6 3s 2

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d w 4p 6 5s 2

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2
ls 2 2s 2 2p 6 3s 2 3p (i 4s 2 3d 10 4p 6 5s 2 4d l0 5p (i 6s 2 4f w 5d l0 6p 6 7s 2

For Group 2A metals, the same rule applies! Magnesium {Mg) is found in row 3 of the Periodic Table, and
its valence electrons are found in the n = 3 energy level. Similarly, Radium {Ra) is found in row 7 of the
Periodic Table, and its valence electrons are found in the n = 7 energy level.

So far so good - but does the same pattern apply to the Group 3A - 8A elements (also known as Groups
13-18). Let's find out by writing the electron configuration for the Group 3A element in each row.

row
row
row
row
row

2
3
4
5
6

B:
Al:
Ga
In :
Tl:

ls 2 2s 2 2p 1

ls 2 2s 2 2p 6 3s 2 3p l

ls 2 2s 2 2p Q 3s 2 3p^As 2 3d w Ap l

ls 2 2s 2 2p 6 3s 2 3p fi 4s 2 3d 10 4p 6 5s 2 4d 10 5p 1

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d w 5p 6 6s 2 5d w 6p 1

Even though the valence electrons in Group 3A elements are found in both s and p orbitals, it turns out
that an element's row still corresponds to the energy level of that element's valence electrons. For example,
Gallium {Ga) is found in row 4 of the Periodic Table, and its valence electrons are found in the n = 4
energy level. Likewise, Thallium {Tl) is found in row 6 of the Periodic Table, and its valence electrons are
found in the n = 6 energy level.

It really does seem as if we can predict the energy level of an element's valence electrons using the row
number for that element in the Periodic Table. Let's try one last example, though, just to be sure by
writing the electron configuration for the Group 7A element in each row.

265

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row
row
row
row
row

B:
Al:
Ga :
In :
Tl:

ls 2 2s 2 2p 5

ls 2 2s 2 2p 6 3s 2 3p 5

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5

ls 2 2s 2 2p Q 3s 2 3p^s 2 3d w Ap^s 2 Ad l %p''

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6 S 2 5d 10 6p 5

Once again, an element's row can be used to determine the energy level of that element's valence electrons.
Chlorine (CI), for instance, is found in row 3 of the Periodic Table, and its valence electrons are found
in the n = 3 energy level. Similarly, Iodine (/) is found in row 5 of the Periodic Table, and its valence
electrons are found in the n = 5 energy level.

You should make note of one final point when it comes to energy levels and how they relate to the Periodic
Table. Our rule for determining the energy level of an element's valence electrons using the element's row
in the Periodic Table works for Group 1A - 8A elements. This rule doesn't apply, however, to the Group
IB - 8B (also known as Groups 3 - 12) elements. The elements in that lower portion of the Periodic Table
(the middle portion of the Periodic Table 'blacked out' in Figure 1) are known as transition metals. They
behave differently, and you can't apply the same rules to them as far as valence electrons are concerned.
The same goes for the isolated lower portion of the Periodic Table (also 'blacked out' in the figure of the
Periodic Table above). This block contains elements known as lanthanides and actinides. Like transition
metals, lanthanides and actinides do not obey the same rules as the Group 1A - 8A elements when it comes
to valence electrons and valence electron energy levels.

Hydrogen and Helium Occupy the First Period

You probably noticed that, in the last section, we didn't mention the first row at all. Instead, we always
started from row 2. The first row in the Periodic Table is a 'special row' for several reasons. To begin with,
the first row of the Periodic Table contains only two elements - hydrogen and helium. Can you figure out
why there are only two elements in the first row?

According to what we learned in the last section, an element's row number corresponds to the energy level
of that element's valence electrons. Therefore, the first row must contain elements with valence electrons
in the n = 1 energy level. By now you should know that there is only one orbital in the n = 1 energy level.
That orbital, of course, is the Is orbital. Hydrogen has one valence electron in the Is orbital, (its electron
configuration is Is 1 ), and helium has two valence electrons in the Is orbital (its electron configuration is
Is 2 ). Since it's impossible to fit more than two electrons into the Is orbital, atoms with a total of three or
more electrons must have valence electrons in an energy level with n = 2 or greater. Clearly, then, atoms
with a total of three or more electrons do not belong in the first row of the Periodic Table.

The first row of the Periodic Table is also special because its elements have special properties. Hydrogen,
for example, is not a metal like the rest of the Group 1A elements. Instead, hydrogen atoms react with each
other and form what's known as hydrogen gas, H<i- As was mentioned in the last lesson, some scientists
will put hydrogen in a category all by itself, rather than including it at the top of the 1A column. We
won't do that here, but you should always keep in mind the fact that hydrogen is 'different,' and that you
shouldn't compare the hydrogen's chemical properties with the chemical properties of the other Group 1A
elements.

Helium is also a special atom. You might wonder why helium appears at the far right-hand side of the
Periodic Table, rather than right next to hydrogen. Again, helium's placement in the Periodic Table reflects
its special chemical properties. Earlier you learned that Group 8A elements were 'inert' and that includes

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266

helium. Even though helium only has two valence electrons, while the rest of the Group 8A elements have
eight valence electrons, helium is placed on the top of the 8A column since helium's chemical behavior is
similar to the chemical behavior of the other Noble Gases because it has a completed outer energy level.

The s Sublevel Block on PT

So far we know that, with the exception of transition metals, lanthanides and actinides, we can use the
row in which an element is found to determine the energy level of that element's valence electrons. Can
the organization of the Periodic Table, and the placement of an element within the Periodic Table, tell us
anything else about the elements electron configuration? The answer is - 'y es -' Remember that the highest
energy valence electrons in Group 1A and Group 2A elements are always in s orbitals. In fact, the only
valence electrons in Group 1A and Group 2A elements are in s orbitals! Lithium, (Li) for instance, has
the electron configuration ls 2 2s 1 . Notice that lithium's single valence electron is in an s orbital. Similarly,
magnesium (Mg) has the electron configuration ls 2 2s 2 2p^'&s 2 . Again, notice that magnesium's two valence
electrons are in an s orbital. Since all of the valence electrons in Group 1A and Group 2A elements exist
in s orbitals, the first two columns of the Periodic Table (columns 1A and 2A) are known as the 's
sublevel block.' The s sublevel block is shown in the figure below. Notice that the s sublevel block consists
of all of the metals from Li down to Fr in column 1A, and all of the metals from Be down to Ra in column
2A. Hydrogen is not included in the s sublevel block, again, because of its special properties.

As pictured below, the s sublevel block of the Periodic Table includes the Group 1A metals and the Group
2A metals. (Source: CK-12 Foundation. CC-BY-SA)

JJ.

BHM.JUM

S Block
/

',.?,_,

'_H

"a;

1LUMIKUM

"Si

"ci

CH1UHFNL

"li

"Cr

CHROMIUM

UlNGlNir.C

"nj

"Se

"zr

"Tc

4 Ru

"Pd

k i

"Cd

"Sb

T e

G1F1IUM

La-Lu

FIUUCIUM

Ac-Lr

if.

Nit

Gut

Unci

Uup

Uuh

Uus

Uuo

The p Sublevel Block on PT

What can we say about the valence electrons in Group 3A- Group 8A elements? In particular, what can
we say about the highest energy valence electrons? If you look carefully, you'll notice that for Group 3A
Group 8A (or if you prefer, Groups 13-18) elements, the highest energy valence electrons are always in p
orbitals. Boron, (B) for instance, has the electron configuration ls 2 2s 2 2p 1 . While boron has both 2s and
2p valence electrons, the 2p valence electrons are higher in energy. Similarly, Krypton (Kr) has the electron
configuration ls 2 2s 2 2p^?>s 2 ?>p & As 2 ?>d w Ap & . Again, while krypton has both 4s and \p valence electrons, the
4p valence electrons are higher in energy. Since the highest energy valence electrons in Group 3A Group
8A elements exist in p orbitals, the final six columns of the Periodic Table (columns 3A through 8A)
are known as the 'p sublevel block.' The p sublevel block is shown in the figure below. Additionally,

267

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as illustrated in the figure below the p sublevel block consists of all of the elements from B down to 77 in
column 3A, all of the elements from C down to Pb in column 4A, all of the elements from TV down to Bi
in column 5A, all of the elements from O down to Po in column 6A, all of the elements from F down to At
in column 7A and, finally, all of the elements from Ne down to Rn in column 8A. Helium is not included
in the p sublevel block, which should make sense, since helium has no p electrons!

As illustrated below, the p sublevel block of the Periodic Table includes the Group 3A - Group 8A elements.
(Source: CK-12 Foundation. CC-BY-SA)

'Li

P Block
\

JL.

FMOiPHUBUi

CNLUBINE

TITtMUM

CHBOMIUM

M1NG1KESE

"nj

"Sr

"Zr

"Sb

",!,.

La-Lu

"la

"lr

"ft

"ti

"Pb

"aj

"Vr

Ac-Lr

Uut

Unci

Uup

Uuh

Uus

Uuo

LANTHANIDES

"Pr

SIHWIUM

T b

"Lu

ACTIHIDES

T h .

UFUWUM

"Pu

"Cf

"Es

Just as the Periodic Table has an s sublevel block and a p sublevel block, it also has a d sublevel block and
an / sublevel block. Defining valence electrons in the d and / sublevel blocks can be more difficult but, in
general, most of the high energy valence electrons in the d sublevel block are in d orbitals while most of
the high energy valence electrons in the / sublevel block are in / orbitals.

As illustrated in the figure below, the Periodic Table is divided into the s sublevel block, the p sublevel
block, the d sublevel block, and the / sublevel block. (Source: CK-12 Foundation. CC-BY-SA)

PERIODIC TABLE OF ELEMENTS

'He

S B Io <* P Block

/ \

D Block

Is..

"aj

ULUHHUM

SILICON

"pi

"Ar

"ft

"Fe

"Nj

"Cu

"lr

"Rb

"Sr

*Tc

"Cd

"in

"Sn

*Sb

",!.,

"Cs

La-Lu

"la

"lr

"Pb

"At

"Ra

Ac-Lr

Uut

Uuq

Uup

Uuh

Uus

Uuo

LANTHANIDES

"Ce

*Eu

EURDPUIH

*Gd

'*Er

YTTEBBIUM

ACTINIDES

T h

"Cf

"Es

A complete summary of the Periodic Table, and the information it contains about an element's valence elec-
trons is shown in the figure below. (Source: http://en.wikipedia.Org/wiki/File:PTable_structure.
png. GNU-FDL)

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268

MAIN GROUP ELEMENTS

Earlier you learned that the Periodic Table was a convenient way to summarize all of the information that
scientists know about the different elements found in our world. The Periodic Table probably even looked
funny to you, because you had no way of understanding what its shape and organization meant. Now that
we've discussed electron orbitals and electron energy levels, though, the Periodic Table shouldn't seem so
strange anymore. In fact, the shape of the Periodic Table actually reflects the way in which electrons are
organized in atoms of the different elements.

The following webpage, Periodic Table (http://www.iun.edu/~cpanhd/periodictable.html) contains
an interactive periodic table that allows you to click on the element and the web page then shows a number
of physical constants for that element. The data includes, name, density, electron configuration, heat of
vaporization, heat of fusion, melting point, boiling point, specific heat, and electronegativity.

Lesson Summary

• For Group 1A, 2A, 3A, 4A, 5A, 6A, 7A and 8A elements, the row in which an element is found
corresponds to the energy level of that element's valence electrons. For example, lithium (Li) is in
row 2, and its valence electrons are in the n = 2 energy level.

• You can predict the energy level of an element's valence electrons using the element's row number in
the Periodic Table.

• Elements in the first row have special properties.

• Hydrogen is not an alkali metal, and is usually found as a gas.

• Helium is a Noble Gas, and exhibits chemical properties similar to the other Noble Gases found in
Group 8A.

• The Periodic Table can be divided into s,d,p and / sublevel blocks.

• For elements in the s sublevel block, all valence electrons are found in s orbitals.

• For elements in the p sublevel block, the highest energy valence electrons are found in p orbitals.

Review Questions

1 . Use the Periodic Table to determine the energy level of the valence electrons in each of the following
elements.

(a) B

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(b) Ga

(d) At

(e) He

2. Fill in the blanks:

(a) B is in the level block of the Periodic Table

(b) Sr is in the level block of the Periodic Table

(d) Cs is in the level block of the Periodic Table

(e) O is in the level block of the Periodic Table

3. Use the Periodic Table to determine the energy level and sublevel of the highest energy electrons in
each of the following elements:

(a)

(b)

(c)

(d)

(e)

4. Decide whether each of the following statements is true or false.

(a) Li has valence electrons in the n = 1 energy level.

(b) Si has valence electrons in the n = 3 energy level.

(d) Xe has valence electrons in the n = 5 energy level.

(e) P has valence electrons in the n = 2 energy level.

5. Match the element to the sublevel block it is found in:

(a) C - i. s sublevel block

(b) Cs - ii. p sublevel block

(d) Cr - iv. / sublevel block

6. The first row of the Periodic Table has:

(a) 1 element

(b) 2 elements

(d) 4 elements

(e) 5 elements

7. Use the Periodic Table to determine which of the following elements has the highest energy valence
electrons.

(a)

(b)

(c)

(d)

(e)

8. Use the Periodic Table to determine which of the following elements has the lowest energy valence
electrons.

(a) Ga

(b) B

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(d) Bi

(e) CI

9. Which energy level does the first row in the d sublevel block correspond to?

Further Reading / Supplemental Links

• http : //learner . org/resources/series61 . html

• http : //dl . clackamas . cc . or . us/chl04-06/valence_electrons . htm

• http : //chemed . chem . purdue . edu/genchem/topicreview/bp/ch8/index . php

• http : //www . cem . msu . edu/~reusch/VirtTxt Jml/intro2 . htm

• http : //en . wikipedia . org/wiki/Valence_electrons

Vocabulary

transition metals Elements in the d sublevel block (columns 1-B through 86) of the Periodic Table.
The highest energy electrons in transition metals are found in d orbitals.

lanthanides and actinides Elements in the / sublevel block of the Periodic Table. The highest energy
electrons in lanthanides and actinides are found in / orbitals.

s sublevel block The elements in the Periodic Table in columns 1A and 2A (excluding hydrogen). All
valence electrons for elements in the s sublevel block are in s orbitals.

p sublevel block The elements in the Periodic Table in columns 3A through 8A (excluding helium).
The highest energy valence electrons for elements in the p sublevel block are in p orbitals.

d sublevel block The elements in the Periodic Table in columns IB through 86 (also known as transition
metals).

/ sublevel block The elements in the lanthanide and actinide rows of the Periodic Table.

noble gases Group 8A elements. These are elements found in the eight column of the Periodic Table.
They are inert, which means that they are very non-reactive.

inert Non-reactive.

Image Sources

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Chapter 9

Relationships Between the
Elements

9.1 Families on Periodic Table

Lesson Objectives

Describe the patterns that exist in the electron configurations for the main group elements.
Identify the columns in the periodic table that contain 1) the alkali metals, 2) the alkaline earth
metals, 3) the halogens, and 4) the noble gases, and describe the differences between each family's
electron configuration.

Given the outermost energy level electron configuration for an element, determine its family on the
periodic table.

Introduction

With the introduction of electron configurations, we began to get a deeper understanding of the periodic
table. An understanding of these electron configurations will prove to be invaluable as we look at bonding
and chemical reactions. The orbital representation method for representing electron configuration is shown
below. The orbital representation was learned in an earlier chapter but like many of the skills you learn in
chemistry, it will be used a great deal in this chapter and in several chapters later in the course.

www.ckl2.org 272

-F

In this lesson, we will focus on the connection between the electron configuration and the main group
elements of the periodic table. We will need to remember the sub-level filling groups in the periodic table.
Keep the following figure in mind. We will use it for the next two chapters.

(Source: Richard Parsons. CC-BY-SA)

Vertical Columns = "groups" or "families"

Horizontal Rows = "Periods"

3A 4A5AGA

J s block

] p block

□ d block

I ] f block

Alkali Metals Have One Electron in Their Outer Energy Level

Elements Ending with s 1 = Alkali

In the periodic table, the elements are arranged in order of increasing atomic number. In previous material
we learned that the atomic number is the number of protons in the nucleus of an atom. For a neutral atom,
the number of protons is equal to the number of electrons. Therefore, for neutral atoms, the periodic table
is also arranged in order of increasing number of electrons. Take a look now at the first group or column

273

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in the periodic table. It is the one marked "1A" in the Period Table figure below. The groups or families
are the vertical rows of elements. The first group has seven elements representing the seven periods of the
periodic table. Remember that a period in the periodic table is a horizontal row. Group 1A is the only
group with seven elements in it.

Group 1A of the Periodic Table (Source: Richard Parsons. CC-BY-SA)

3 A 4A5A6A

s block

p block

Lj d block

□ f block

Table 9.1: Electron Configurations for Group 1 Metals

Element

Atomic Number

Electron Configuration

Lithium (Li)
Sodium (Na)
Potassium (K)
Rubidium (Rb)
Cesium (Cs)

Francium (Fr)

3
11
19
37

ls 2 2s 2 2p 6 3s x

ls 2 2s 2 2p 6 3s 2 3p 6 4s 1

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d w 4p 6 5s 2 4d 10

5p 6 6^

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10

5p 6 6s 2 4f u 5d 10 6p 6 7s 1

What do you notice about all of the elements in Group 1? They all have s 1 as the outermost energy level
electron configuration. The whole number in front of the "s" tells you what period the element is in. For
example sodium, Na, has the electron configuration ls 2 2s 2 2p 6 3s 1 , so it is in period 3. It is the first element
of this period.

This group of elements is called the alkali metals. They get their name from ancient Arabic (al kali)
because "scientists" of the time found that the ashes of the vegetation they were burning contained a large
amount of sodium and potassium. In Arabic, al kali means ashes. We know today that all alkali metals
have electronic configurations ending in s 1 . You might want to note that while hydrogen is often placed in
group 1, it is not considered an alkali metal. The reason for this will be discussed later.

Alkaline Earth Elements Have Two Electrons in Their Outer En-
ergy Level

Elements Ending with s 2 — Alkaline Earth

Taking a look at Group 2 A in Table 9.2, we can use the same analysis we used with group 1 to see if

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274

we can find a similar trend. It is the second vertical group in the periodic table and it contains only six
elements.

What do you notice about all of the elements in group 2A? They all have an outermost energy level electron
configuration of s 2 . The whole number in front of the "s" tells you what period the element is in. For
example, magnesium, Mg, has the electron configuration ls 2 2s 2 2p e 3s 2 , so it is in period 3 and is the second
element in that period. Remember that the s sublevel may hold two electrons, so in Group 2A, the s orbital
has been filled.

Elements in this group are given the name alkaline earth metals. They get their name because early
"scientists" found that all of the alkaline earths were found in the earth's crust. Alkaline earth metals,
although not as reactive as the alkali metals, are still highly reactive. All alkaline earth metals have
electron configurations ending in s 2 .

Table 9.2: Electronic Configurations for Group 2A Metals

Element

Atomic Number

Electron Configuration

Beryllium (Be)
Magnesium (Mg)
Calcium (Ca)
Strontium (Sr)
Barium (Ba)

Radium (Ra)

12
20
38
56

ls 2 2s 2

ls 2 2s 2 2p^s 2

ls 2 2s 2 2p 6 3s 2 3/? 6 4s 2

ls 2 2s 2 2p & 2,s 2 2,p & As 2 ?,d w Ap & hs 2

ls 2 2s 2 2p 6 3s 2 3p 6 As 2 3d w Ap & hs 2 Ad w

hp%s 2

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4/? 6 5s 2 4d 10

5p%s 2 4f u 5d 10 6p 6 7s 2

Noble Gases Have 8 Electrons in Their Outer Energy Level

Elements Ending with s 2 p® = Noble Gases

(Source: Richard Parsons. CC-BY-SA)

3A 4A 5A &A

s block
p block

□ d block

□ f block

The first person to isolate a noble gas was Henry Cavendish, who isolated argon in the late 1700 s. The
noble gases were actually considered inert gases until the 1960s when a compound was formed between
xenon and fluorine which changed the way chemists viewed the "inert" gases. In the English language,
inert means to be lifeless or motionless; in the chemical world, inert means does not react. Later, the name
"noble gas" replaced "inert gas" for the name of Group 8A.

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When we write the electron configurations for these elements, we see the same general trend that was
observed with groups 1A and 2A; that is, similar electron configurations within the group.

Table 9.3: Electron Configurations for Group 8A Gases

Element

Atomic Number

Electron Configuration

Helium (He)
Neon (Ne)
Argon (Ar)
Krypton (Kr)
Xenon (Xe)

Radon (Rn)

2
10

28
36
54

ls z

ls 2 2s 2 2p 6

ls 2 2s 2 2p 6 3s 2 3p 6

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10

5p e

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10

Bp 6 6s 2 4f u 5d 10 6p 6

Aside from helium, He, all of the noble gases have outer energy level electron configurations that are the
same,ns 2 np & , where n is the period number. So Argon, Ar, is in period 3, is a noble gas, and would therefore
have an outer energy level electron configuration of 3s 2 3p 6 . Notice that both the s and p sublevels are
filled. Helium has an electron configuration that might fit into Group 2A. However, the chemical reactivity
of helium, because it has a full first energy level, is similar to that of the noble gases.

Halogens Have 7 Electrons in Their Outer Energy Level

Elements Ending with s 2 p 5 = Halogens
(Source: Richard Parsons. CC-BY-SA)

3A 4A 5A SA

s block
p block

□ d block

□ f block

The halogens are an interesting group. Halogens are members of Group 7A, which is also referred to as
17. It is the only group in the periodic table that contains all of the states of matter at room temperature.
Fluorine, F2, is a gas, as is chlorine, Ch- Bromine, Br2, is a liquid and iodine, I2, and astatine, At2, are both
solids. What else is neat about Group 7A is that it houses four (4) of the seven (7) diatomic compounds.
Remember the diatomics are H2,N2,02,F2,Ch, Br?, and h- Notice that the latter four are Group 17
elements. The word halogen comes from the Greek meaning salt forming. French chemists discovered
that the majority of halogen ions will form salts when combined with metals. We all know some of these
already: LiF,NaCl,KBr, and Nal.

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276

Taking a look at Group 7A in the figure, we can find the same pattern of similar electron configurations
as found with group 1A, 2A, and 8A. It is the 17" 1 group in the periodic table and it contains only five
elements.

Table 9.4: Electron Configurations for Group 7A Elements

Element

Atomic number

Electronic configuration

Fluorine (F)
Chlorine (CI)
Bromine (Br)
Iodine (/)

Astatine (At)

ls 2 2s 2 2p 5

ls 2 2s 2 2p 6 3s 2 3p 5

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10

5p 5

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10

5p 6 6s 2 4f M 5d 10 6p 5

What is the general trend for the elements in Group 7A? They all have, as the outermost energy level
electron configuration, ns 2 np 5 , where n is the period number. You should also note that these elements
are one group away from the noble gases (the ones that generally don't react!) and the outermost electron
configuration of the halogens is one away from being filled. For example, chlorine (CI) has the electron
configuration [Ne] 3s 2 3p 5 so it is in period 3, the seventh element in the main group elements. The
main group elements, as you recall, are equivalent to the s + p blocks of the periodic table (or the teal and
purple groups in the diagram above).

The Oxygen Family Has 6 Electrons in the Outer Energy Level

Elements Ending with s 2 p 4 = the Oxygen Family

(Source: Richard Parsons. CC-BY-SA)

3A 4A 5A &A

s block
p block

□ d block

□ f block

Oxygen and the other elements in Group 6 A have a similar trend in their electron configurations. Oxygen
is the only gas in the group; all others are in the solid state at room temperature. Oxygen was first named
by Antoine Lavoisier in the late 1700 s but really the planet has had oxygen around since plants were first
on the earth.

Taking a look at Group 6A in the figure below, we find the same pattern in electron configurations that we
found with the other groups. Oxygen and its family members are in the 16 f/l group in the periodic table.

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In Group 16, there are, again, only five elements.

Table 9.5: Electron Configurations for Group 6A Elements

Element

Atomic Number

Electron Configuration

Oxygen (O)
Sulfur (S)
Selenium (Se)
Tellurium (Te)

Polonium (Po)

34
52

ls 2 2s 2 2p A

ls 2 2s 2 2p^s 2 ?,p±

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p' i

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d w 4p 6 5s 2 4d w

5/? 4

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10

5p 6 6s 2 4f 14 5d 10 6p' i

When we examine the electron configurations of the Group 6A elements, we see that all of these ele-
ments have the outer energy level electron configuration of ns 2 np A . We will see that this similar electron
configuration gives all elements in the group similar properties for bonding.

These elements are two groups away from the noble gases and the outermost electron configuration is two
away from being filled. Sulfur, for example, has the electron configuration ls 2 2s 2 2/? 6 3s 2 3/? 4 so it is in
period 3. Sulfur is the sixth element in the main group elements. We know it is the sixth element across
the period of the main group elements because there are 6 electrons in the outermost energy level.

The Nitrogen Family Has 5 Electrons in the Outer Energy Level

Elements Ending with s 2 p 3 = the Nitrogen Family

(Source: Richard Parsons. CC-BY-SA)

3A 4A 5A &A

s block
p block

□ d block

□ f block

Just as we saw with Group 16, Group 5A has a similar oddity in its group. Nitrogen is the only gas in
the group with all other members in the solid state at room temperature. Nitrogen was first discovered
by the Scottish chemist Rutherford in the late 1700' s. The air is mostly made of nitrogen. Nitrogen has
properties that are different in some ways from its group members. As we will learn in later lessons, the
electron configuration for nitrogen provides the ability to form very strong triple bonds.

Nitrogen and its family members belong in the 15 group in the periodic table. In Group 15, there are

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278

also only five elements.

Table 9.6: Electronic Configurations for Group 5A Elements

Element

Atomic Number

Electron Configuration

Nitrogen (N)
Phosphorus (P)
Arsenic (As)
Antimony (Sb)

Bismuth (Bi)

15
33
51

ls 2 2s 2 2p 3

ls 2 2s 2 2p & 3s 2 3p 3

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 3

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d w 4p 6 5s 2 4d 10

5p 3

ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10

5p 6 6s 2 4f 14 5d 10 6p 3

What is the general trend for the elements in group 5A? They all have, as the electron configuration in
the outermost energy level, ns 2 np 3 , where n is the period number. These elements are three groups away
from the noble gases and the outermost energy level electron configuration is three away from having a
completed outer energy level. In other words, the p sublevel in the Group 15 elements is half full. Arsenic,
for example, has the electron configuration ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 3 so it is in period 4, the fifth element
in the main group elements. We know it is the fifth element across the period of the main group elements
because there are 5 electrons in the outermost energy level.

Lesson Summary

• Families in the periodic table are the vertical columns and are also referred to as groups.

• Group 1A elements are the alkali metals and all have one electron in the outermost energy level
because their electron configuration ends in s .

• Group 2A elements are the alkaline earth metals and all have two electrons in the outermost energy
level because their electron configuration ends in s 2 .

• Group 5A elements all have five electrons in the outermost energy level because their electron con-
figuration ends in s 2 p 3 .

• Group 6A elements all have six electrons in the outermost energy level because their electron config-
uration ends in s 2 p 4 .

• Group 7A elements are the halogens and all have seven electrons in the outermost energy level because
their electron configuration ends in s 2 p 5 .

• Group 8A elements are the noble gases and all have eight electrons in the outermost energy level
because their electron configuration ends in s 2 p®.

• Elements in group 8A have the most stable electron configuration in the outermost shell because the
sublevels are completely filled with electrons.

Review Questions

1. If an element is said to have an outermost electronic configuration of ns 2 np 3 , it is in what group in
the periodic table?

(a) Group 3A

(b) Group 4A

(d) Group 7A

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2. What is the general electronic configuration for the Group 8A elements? (Note: when we wish to indi-
cate an electron configuration without specifying the exact energy level, we use the letter "»" to repre-
sent any energy level number. That is, ns 2 np s represents any of the following; 2s 2 2p s , 3s 2 3p s , 4s 2 4/? 3 ,
and so on.)

(a) ns 2 np 6

(b) ns 2 np 5

(d) ns 2

3. The group 2 elements are given what name?

(a) alkali metals

(b) alkaline earth metals

(d) noble gases

Using the diagram above, iden-

T c

■

tify:

(a) The alkali metal by giving the letter that indicates where the element would be located and
write the outermost electronic configuration.

(b) The alkaline earth metal by giving the letter that indicates where the element would be located
and write the outermost electronic configuration.

(c) The noble gas by giving the letter that indicates where the element would be located and write
the outermost electronic configuration.

(d) The halogen by giving the letter that indicates where the element would be located and write
the outermost electronic configuration.

(e) The element with an outermost electronic configuration of s 2 p s by giving the letter that indicates
where the element would be located.

(f) The element with an outermost electronic configuration of s 2 p 1 by giving the letter that indicates
where the element would be located.

5. In the periodic table, name the element whose outermost electronic configuration is found below.
Where possible, give the name of the group.

(a) 5s 2

(b) 4s 2 3d 10 4p 1

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280

(d) 5s 2 4d 10 5p 2

(e) 3s l

(f) l, 2

(g) 6s 2 5d 10 6p 5
(h) 4s 2 4p 4

Further Reading / Supplemental Links

• http://en.wikipedia.org

Vocabulary

group columns of the periodic table.

period Horizontal rows of the periodic table.

alkali metals Group 1 in the periodic table {Li,Na,K,Rb,Cs,Fr).

alkaline earth metals Group 2 in the periodic table {Be,Mg,Ca,Sr,Ba,Ra).

noble gases Group 18 in the periodic table (He,Ne,Ar,Kr,Xe,Rn).

halogens Group 17 in the periodic table (F, CI, Br, I, At).

main group elements Equivalent to the s + p blocks of the periodic table, also known as "representative
elements."

9.2 Electron Configurations

Lesson Objectives

• Convert from orbital representation diagrams to electron configuration codes.

• Distinguish between outer energy level (valence) electrons and core electrons.

• Use the shortcut method for writing electron configuration codes for atoms and ions.

Introduction

In the last section, we determined that the members of the chemical families (vertical columns on the
periodic table) all have the same electron configuration in their outermost energy levels. In the next
section, we will review the orbital representation for electron configuration and the electron configuration
code. Then we will learn a shortcut for writing electron configuration codes.

Shortcut for Writing Electron Configurations

The diagram below represents the orbital representation diagram used in earlier chapters. The orbital
representation diagram has a few different variations but all are used to draw electron configurations. Most

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show the orbitals in groups, as lines, boxes, or circles with each orbital having its own line (or circle) within
each sublevel.

o 9°oooooooooooo

^ °Q° ooooo

LJ 4d

5s ooo

o u - ooooo

V_X 3d

X ooo

O 3p

A ooo

O 2p

Is
The sublevel 2p has 3 orbitals (2p x ,2p y , and 2p z ) and each of these orbitals has its own line (or box).

2p* 2 Pl

Let's begin this section with the orbital box (or the orbital representation diagram) for a neutral atom.
Draw the electronic configuration for potassium using the electron configuration diagram below. Remember
that potassium is element number 19 so has 19 electrons. Every line in the energy diagram below holds 2
electrons of opposite spins.

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M
fi
(0
A
4)
I*
O

ooo

°9° ooooo

U> 4d

5s OOO

q u v u ooooo
* ©®©

O 3p

A©©©

Now we can simplify this by excluding all of the levels and sublevels that are not in use or do not hold
electrons.

u
V

c
w

bJO

" ©©©

"©©©
© 2 p

Writing the electron configuration code for potassium, we would get:

K : ls 2 2s 2 2p 6 3s 2 3p 6 4s 1 .

Remember, the noble gases are the elements that are nearly non-reactive, partially due to the fact that the
outermost energy level electron configuration was full (ns 2 np 6 ). Look at potassium on the periodic table
and then find the preceding noble gas. Drag you finger counting backward from potassium to the nearest
noble gas. What do you get? Argon (Ar) is correct. Now, let's draw the electron configuration for Argon
and place the electron configuration for potassium beside it.

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-L

T5 -

H H H

Electron configuration for Argon

H U H

t* t* t*

Electron configuration for Potassium

What do you notice looking at the two diagrams? Potassium has the same electron configuration as argon
+ 1 electron. We can make this point when we write electronic configuration codes.

2o r .2 „6Q 2o»,6 / |„l

K : ls z 2s z 2p°3s z 3p°4s

„2o„2

2o„6

Ar : ls z 2s z 2p°3s z 3p

Therefore,

K : [Ar]4j 1 . {This is known as the noble gas configuration code system or the "short-cut"}

Let's try another. Draw and write the electron configuration of gallium (Go). Gallium is number 31 and
therefore has 31 electrons. It is also in the 4 period, the third element in the period of main group
elements. From this information, we know that the outermost electrons will be As 2 Ap l .

3s 3p

It has 3 valence electrons so it
is family 3A (column 13).

In period 3

Electron configuration of Gallium (Ga).

Writing the electron configuration code for gallium, we get: Ga : ls 2 2s 2 2p e 3s 2 3p e 4:S 2 3d l0 4:p l . In numerical
order, the electronic configuration for gallium is: Ga : ls 2 2s 2 2p e 3s 2 3p e 3d 10 4s 2 4:p 1 .

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284

<1.

^ u r+ n n

ti U H

U ti U

Electron configuration for Gallium [Ga;

Now, go back to the previous noble gas. Again, we find the previous noble gas is argon. Remember argon
has electronic configuration of ls 2 2s 2 2p e 3s 2 3p e .

Ga : ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 1

Ar : ls 2 2s 2 2p 6 3s 2 3p (:

Therefore,

Ga: [Ar]4s 2 3d 10 V.

This seems to work well for neutral atoms, what about ions? Remember that a cation is formed when
electrons are removed from a neutral atom and anions are formed when electrons are added to a neutral
atom. But where do the electrons get added to or removed from? The electrons that are involved in
chemical reactions are the ones that are in the outer shell. Therefore, if electrons are to be removed or
added, they get removed from or added to the shell with the highest value of n.

Let's now look at an example. Draw the electron configuration for Fe. Then, write the reaction for the
formation of Fe 2+ .

285

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Els

H H H

UJ_i_i_i

Electron configuration for Iron (Fe)

The reaction for the formation of the Fe 2+ ion is: Fe — > Fe 2+ + 2e~. Therefore we have to remove 2
electrons. From where do we remove the two electrons? The rule is to remove from the highest n value
first which, in this case, means we remove from the electrons from the 4 s sublevel first.

Fe -» Fe 2+ + 2e

„2o„2r

„2qj6

2 0o 2r, 6q„2o 6q j6

ls'2s'2p 3s'3p°4s'3cF -» ls'2s'2p 3s'3p°3d° + 2e

Remember that when we write electron configuration codes using noble gas configurations we use the
previous noble gas. In writing the electron configuration code for Fe or Fe 2+ , we would use the symbol
[Ar] to substitute for ls 2 2s 2 2p 6 3s 2 3p 6 .

Fe -» Fe 2+ + 2e~
[Ar]3d 6 4s 2 -» [Ar]3J 6 + 2e~

In addition to forming Fe 2+ ions, iron atoms are also capable of forming Fe 3+ ions. Extra stability is gained
by electron configurations that have exactly half-filled d orbitals. The third electron that is removed from
an iron atom to form the Fe s+ ion is removed from the d orbital giving this iron ion a d orbital that is
exactly half-filled. The extra stability of this configuration explains why iron can form two different ions.

Fe -» Fe 3+ + 3e

[Ai]3ds 6 4:s 2 -» [Ar]3ds 5 + 3e~

Now what about negative ions, the anions? Remember that negative ions are formed by the addition of
electrons to the electron cloud. Look at bromine, Br. Neutral bromine has atomic number 35 and therefore,
holds 35 electrons.

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n n t

L f

w u n n w

& iii
u

Jilt it

it
ts

Electron configuration for Bromine (Br)

The reaction for the formation of the Br~ ion is: Br + e~ —* Br~. The bromide ion is formed by the addition
of one electron. If you look at the electron configuration for bromine, you will notice that the 4p sublevel
needs one more electron to fill the sublevel and if this sublevel is filled, the electron configuration for Br is
the same as that of the noble gas krypton (Kr). When adding electrons to form negative ions, the electrons
are added to the lowest unfilled n value.

Br + e
ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5 + e

Br~

ls 2 2s 2 2/3s 2 3/4s 2 3d 10 4/7 e

Br + e~ -» Br
ls 2 2s 2 2p^s 2 ?,p (i As 2 M w Ap h + e~ -» [Kr]

Notice how Br~ has the same electron configuration as Kr. When this happens, the two are said to be
isoelectronic.

Sample Question: Write the electron configuration codes for the following atoms or ions using the noble
gas electron configuration shortcut.

(a) Se

(b) Sr

(d) Ca 2+

(e) Al 3+
Solution:

(a) Se[Ar] : As 2 3d w Ap A

(b) Sr[Kr] : 5s 2

(d) Ca 2+ : [Ar]

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(e) AP+ : [Ne]

Core Electrons

1 2A
2

3A 4A 5A 6A 7 a
13 14 15 16 17

8A
18

3 4 5 6 7 8 9 10 11 12

When we look at the electron configuration for any element we can distinguish that there are two different
types of electrons. There are electrons that work to do the reacting and there are those that don't. The
outer shell electrons are the valence electrons and these are the electrons that are responsible for the
reactions that the atoms undergo. The number of valence electrons increases as you move across the
periodic table for the main group elements.

In older periodic tables, the groups were numbered using an A and B group numbering system. The
representative groups (s and p blocks) were numbered with A and the transition elements were numbered
with B. The table on the right has both numbering systems. The A numbering is shown for you in the
diagram in red; this is all we will be concerned with for this lesson. The number of valence electrons in the
A groups increases as the group number increases. In fact, the number of the A groups is the number of
valence electrons. Therefore, if an element is in group 3A, it has 3 valence electrons, if it is in group 5A,
it has 5 valence electrons.

All electrons other than valence electrons are called core electrons. These electrons are in inner energy
levels. It would take a very large amount of energy to pull an electron away from these full shells, hence,
they do not normally take part in any reactions. Look at the electron configuration for gallium (Go). It
has 3 valence electrons and 28 core electrons.

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288

<1.

^ u r+ n n

ti U H

U ti U

Electron configuration for Gallium [Ga;

Sample Question: How many core electrons and valence electrons are in each of the following?

(a) V

(b) As

(d) Xe
Solution:

(a) 23 y '■ Core Electrons = 18, Valence electrons = 5
(b)33 As : Core Electrons = 28, Valence electrons = 5

(d) 54 Xe : Core Electrons = 46, Valence electrons = 8

Lesson Summary

• An orbital box diagram can be used to draw electron configurations for elements.

• The noble gas configuration system is the shorthand method for representing the electron configura-
tion of an element.

• The noble gas symbol represents all of, or most of, the core electrons.

• For cations, the electrons are removed from the outermost sublevel having the greatest n value.

• For anions, the electrons are added to the unfilled energy level having the lowest n value.

• Isoelectronic species have the same electron configurations.

• Using an orbital box diagram, the core electrons and the valence electrons can easily be visualized.

• The valence electrons are those electrons that are in the outermost shell.

Review Questions

1. What is the difference between the standard electron configuration code and the electron configuration
code using noble gas configurations?

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2. The standard electron configuration is often called the ground state electron configuration. Why do
you think this is so?

3. How is it possible to have the same number of core electrons and valence electrons for two different
ions?

4. Draw the orbital representation for the electron configuration of potassium, K.

5. Write the electron configuration code for potassium, K.

6. Write the noble gas electron configuration code for potassium, K.

7. How many core electrons and valence electrons are in potassium, Kl

8. Write the electron configuration for potassium, K + .

9. Write the noble gas electron configuration code for potassium, K + .

10. With what species is K + isoelectronic?

11. The electron configuration below is for which element^ 2

(a)

(b)

(c)

(d)

. The electron configuration below is

for whi^ e eLe i njg 1 t 2 ? 6/? 4

(a)

(b)

(c)

(d)

. What is the noble gas electron configuration for the bromine element?

(a)

ls 2 2s 2 2p 6 3s 2 3p 6 4:S 2 4:p 5

(b)

ls 2 2s 2 2p 6 3s 2 3p 6 3d l0 4s 2 4p 5

(c)

[Ai]4s 2 4p 5

(d)

[Ai]3d 10 4s 2 4p 5

. Write the noble gas electron configuration for each of the following.

(a)

(b)

N 3 -

(c)

Sr 2 +

(d)

Sn 2+

(e)

. How

many core electrons and valence electrons are in each of the following?

(a) Mg

(b)

(c)

(d)

(e)

Vocabulary

orbital box diagram A diagram for drawing the electron configurations where sub-levels are shown in
groups (or even in boxes) and each orbital has its own line (or box) within each sub-level.

isoelectronic Having the same electron configuration.

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core electrons Electrons that occupy energy levels below the outermost energy level.
valence electrons Electrons that occupy the outer shell of the atom or ion.

9.3 Lewis Electron Dot Diagrams

Lesson Objectives

• Explain the meaning of an electron dot diagram.

• Draw electron dot diagrams for given elements.

• Describe the patterns of electron dot diagrams in the periodic table.

Introduction

This chapter will explore yet another shorthand method of representing the valence electrons. The method
explored in this lesson will be a visual representation of the valence electrons. We will, as we observed
in the previous lesson, finish the lesson with a look at how this visual representation flows in a pattern
throughout the periodic table.

A Simplified Way to Show Valence Electrons

As defined earlier in this chapter, core electrons are all the electrons except the valence electrons and valence
electrons are the electrons in the outermost energy level. Valence electrons are the electrons responsible
for chemical reactions. Here is the electron configuration for sodium.

i
■11

3s-

u u u

"sr

Electron configuration for Sodium (Na)

The electron configuration is Na : ls 2 2s 2 2p e 3s 1 . The core electrons are ls 2 2s 2 2p®. The valence electron
is 3s 1 . One way to represent this valence electron, visually, was developed by Gilbert N. Lewis. These
visual representations were given the name Lewis electron dot diagrams. Lewis suggested that since
the valence electrons are responsible for chemical reactions and the core electrons are not involved, we
should use a diagram that shows just the valence electrons for an atom.

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To draw a Lewis electron dot diagram for sodium you can picture the symbol for sodium in a box with
the box having four sides. Each side of the box represents either the s or one of the three p orbitals in
the outermost energy level of the atom. The first and second valence electrons are placed on the side
representing the s orbital and the next electrons are placed in the p orbitals. Just as the electrons are
placed singly before being doubled up in the orbital representation, so they are placed one at a time in the
p orbitals of an electron dot formula. A single dot represents one valence electron. Thus, the Lewis dot
formula for sodium is:

Na*

Look at the electron configuration for magnesium. Magnesium is element #12.

b£i

'«

U H U

H
H

Electron configuration for Magnesium (Mg)

To draw the Lewis electron dot diagram we picture in our minds the symbol for Mg in a box with all of its
core electrons (i.e., ls 2 2s 2 2p (i ). Then we place the valence electrons around the sides of the box with each
side representing an orbital in the outermost energy level. How many valence electrons does magnesium
have? Correct, there are 2 valence electrons, 3s 2 .

Therefore the Lewis electron dot formula for magnesium is:

M g :

Look at the electron configuration for chlorine.

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292

H H t

ffl

Electron configuration for Chlorine (CI)

The electron configuration for chlorine could be written as: CI : ls 2 2s 2 2p®3s 2 3p 5 . The core electrons would
be ls 2 2s 2 2p 6 while the valence electrons would be in the third shell (or where n = 3). Therefore, chlorine
has 7 valence electrons. The Lewis electron dot diagram would look like the following:

• •

:ci*

• •

Notice that the electrons are in groups of two. Think of the chlorine in a box and the box has 4 sides.
Each side can have 2 electrons on it. Therefore there can be a maximum of 2 x 4 = 8 electrons normally
on any Lewis electron dot diagram.

Sample question: Write the Lewis electron dot formula for:

(a) Oxygen

(b) Sulfur

(d) Carbon

Solution:

(a) Oxygen has the electron configuration: ls 2 2s 2 2p 4 , therefore there are 2 core electrons and 6 valence
electrons. The Lewis electron dot formula is:

:o*

(b) Sulfur has the electron configuration: ls 2 2s 2 2p®3s 2 3p 4 , therefore there are 10 core electrons and 6
valence electrons. The Lewis electron dot formula is:

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:s *

(c) Potassium has the electron configuration: 1j 2 2j 2 2/? 3j 2 3/? 6 4j , therefore there are 18 core electrons and
1 valence electron. The Lewis electron dot formula is:

(d) Carbon has the electron configuration: ls 2 2s 2 2p 2 , therefore there are 2 core electrons and 4 valence
electrons. The Lewis electron dot formula is:

All the Elements in a Column Have the Same Electron Dot Dia-
gram

In the previous lesson, it was shown that using the earlier version for numbering the periodic table, we
could see a pattern for finding the number of valence electrons in each of the groups in the main group
elements. Since the Lewis electron dot diagrams are based on the number of valence electrons, it would
hold true that the elements in the same group would have the same electron dot diagram. In other words,
if every element in Group 1A has 1 valence electron, then every Lewis electron dot diagram would have
one single dot in their Lewis electron dot diagram. Take a look at the table below.

Table 9.7: Lewis Electron Dot Diagrams for Group 1A Elements

Element # Valence e~ Lewis electron dot diagram

Hydrogen (H) 1

Lithium (Li) 1

Sodium (Na)

Potassium (K)
Rubidium (Rb)

Cesium (Cs)
Francium (Fr)

H*
Li-

Rb*

Cs*

Fr-

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The same pattern holds for all elements in the main group. Look at Table 9.8 for the Lewis dot diagrams
for Group 2 A elements.

Table 9.8: Lewis Electron Dot Diagrams for Group 2A Elements

Element

# Valence e~

Lewis electron dot diagram

Beryllium (Be)
Magnesium (Mg)
Calcium (Ca)
Strontium (Sr)
Barium (Ba)
Radium (Ra)

Be
Mg
Ca
Sr
Ba
Ra

A similar pattern exists for the Lewis electron dot diagrams for Group 3A.

Table 9.9:

Element

# Valence e

Lewis electron dot diagram

Boron (B)
Aluminum (Al)
Gallium (Ga)
Indium (In)

Thallium (Tl)

••
Al*

••
Ga*

••
In-

• •
Tl*

Lesson Summary

• A Lewis Electron Dot Diagram is used to visually represent only the valence electrons.

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• The core electrons are symbolized using the symbol of the element with the valence electron dots
surrounding the element symbol.

• The maximum number of valence electron dots in the Lewis electron dot diagram is 8. Two electrons
can go one each side (top, bottom, left, and right).

• Electrons are added to the electron dot formula by placing the first two valence electrons in the s
orbital, then one in each p orbital until each p orbital has one electron, and then doubling up the
electrons in the p orbitals.

• Each element in a group has the same number of valence electrons and therefore has the same Lewis
electron dot diagram. (Same number of dots, different symbol.)

Review Questions

1. What is the advantage of the Lewis electron dot diagram?

2. What is the maximum number of dots in a Lewis's Electron dot diagram?

3. Draw the Lewis electron dot diagram for lithium.

4. Draw the Lewis electron dot diagram for calcium.

5. Draw the Lewis electron dot diagram for bromine.

6. Draw the Lewis electron dot diagram for selenium.

7. Write the Lewis electron dot diagrams for each of the following. What observations can you make
based of these diagrams?

(a) Oxygen 2- ion

(b) Sulfur 2- ion

(d) Aluminum

8. Write the trend for the Lewis electron dot diagrams for Group 6A (or Group 16) by filling in the
table below.

Table 9.10:

Element # Valence e~ Lewis Electron Dot Diagram

Oxygen (O)
Sulfur (S)
Selenium (Se)
Tellurium (Te)
Polonium (Po)

Further Reading / Supplemental Links

• http://en.wikipedia.org

Vocabulary

Lewis Electron Dot Diagram A shorthand visual representation of the valence electrons for an ele-
ment. (Lewis electron dot diagram for sodium with one valence electron:

Xa"

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9.4 Chemical Family Members Have Similar Prop-
erties

Lesson Objectives

• Explain the role of the core electrons.

• Explain the role of valence electrons in determining chemical properties.

• Explain how the chemical reactivity trend in a chemical family is related to atomic size.

Introduction

Combining what we have learned from the previous lessons, we can now determine how the core electrons
and the valence electrons relate to the properties of families in the periodic table. We have looked at the
trend of the number of valence electrons within each group and now we will expand this to include a look
at the trends that exist between the sublevel of valence electrons and the relative ability of an element to
react in a chemical reaction. We will explore the possibility of predicting chemical reactivity based on the
number of valence electrons and the sublevel to which they belong. Let's begin our lesson with a review of
the core electrons and discuss their application in chemical reactions. Following this, we will have a similar
discussion about valence electrons.

Core Electrons Can be Ignored in Determining an Element's
Chemistry

As learned in an earlier lesson, the core electrons are the inner electrons found in the electron configuration
for the element. These electrons fill up the inner sublevels and are thus not responsible for bonding and
are not involved in chemical reactions. The core electrons also are not directly responsible for determining
the properties of the elements. Remember that the core electrons represent all of the electrons except for
the valence electrons.

Look at the electron configuration for beryllium (Be).

Be : ls 2 2s 2

There are two core electrons (Is 2 ) and two valence electrons (2s 2 ).

When we look at the electron configuration for Selenium (Se), we see the following.

Se : ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4

There are 28 core electrons (ls 2 2s 2 2p e 3s 2 3p®3d 10 ) and six valence electrons (4s 2 4/? 4 ). Again, none of the
core electrons from either of these elements, or any element for that matter, will participate in chemical
reactions.

Sample question: What are the core electrons in each of the following?

(a) : ls 2 2s 2 2p 4

(b) In: [Kr]5s 2 4d 10 5p l
Solution:

(a) O: core electrons = Is 2 , therefore there are two of them.

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(b) In: indium has three valence electrons and all the rest are core electrons. The core electrons include
all the electrons in the energy levels below n = 5. (Ad 10 are core electrons since they are held in a lower
energy level.)

Valence Electrons Determine Chemical Properties

The valence electrons, unlike the core electrons, are responsible for all of the chemical reactions that take
place between elements. They determine the properties of the elements. We have already learned that
all metallic elements in the main group elements are able to lose their valence electrons since the energy
requirements to do so are relatively low. The non-metallic elements (other than noble gases) are able to
gain electrons readily because of high electron affinities. For elements in Group 3, the atoms will have to
lose three electrons to have electron configurations similar to the previous noble gas. For elements in group
6, they will have to gain two electrons to have similar electron configurations as their nearest noble gas.
Earlier we determined that all elements in the same group have the same Lewis Electron Dot diagram.
What this means is that all elements in the same group have the same number of valence electrons and
will react the same way to gain or lose electrons when participating in reactions. The number of valence
electrons determines what types of chemical properties the elements will have.

When you look at Table 9.11, you can see that the Group 1A metals all have the same number of
valence electrons and also have the same appearance. These metals also react the same way under similar
conditions.

Table 9.11: Electron Configuration and General Appearance for Group 1A Elements

Element Electron Configuration Appearance

Li ls 2 2s 1 Silvery-white metal

Na ls 2 2s 2 2p 6 3s 1 Soft, silvery-white metal

K ls 2 2.s 2 2p 6 3.s 2 3p 6 4.s 1 Soft, silvery-white metal

Rb ls 2 2.s 2 2/7 6 3.y 2 3p 6 4.y 2 3rf 10 4/? 6 5.s 1 Soft, silvery-white metal

Cs ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d w 4p 6 5s 2 4d w Soft, silvery-white metal with

5p 6 6i 1 gold sheen

Fr ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d w 4p 6 5s 2 4d w ? too short lived

5p 6 6s 2 4f 14 5d 10 6s 1

We can see that all of the elements in the Group 1A metals all have one valence electron in their outer
energy levels. This means that they can lose this one electron and become isoelectronic with a noble gas
configuration. Thus, elements in Group 1A will readily lose this electron because it takes very low energy
to remove this one outer electron.

As the atomic number increases for the elements in the alkali metal family group, the valence electron is
further away from the nucleus. The attraction between the valence electron and the nucleus decreases as
the atomic size increases. The further the electrons are away from the nucleus, the less hold the nucleus
has on the electrons. Therefore the more readily the electrons can be removed and the faster the reactions
can take place. So, the valence electrons will determine what reactions will occur and how fast they will
occur based on the number of electrons that are in the valence energy level. All group 1A elements will
lose one electron and form a 1+ cation.

Looking at Table 9.12 for the electron configuration of the Group 2A metals, we see that the outer energy
level holds two electrons for each of the metals in group 2.

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Table 9.12: Electron Configuration for Group 2 A Elements

Element Electronic Configuration

Beryllium (Be) ls 2 2s 2

Magnesium (Mg) ls 2 2s 2 2p 6 3s 2

Calcium (Ca) ls 2 2s 2 2p 6 3s 2 3p 6 4s 2

Strontium (Sr) ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 bs 2

Barium (Ba) ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2

Radium (Ra) ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d w 4p 6 5s 2 4d 10 5p 6 6s 2 4f u 5d w 6p 6 7s 2

As with the Group 1A metals, the group 2A metals will form cations by losing the s electrons, but this
time they will lose two electrons. And, as with the Group 1A elements, the elements are more reactive as
the atomic number increases because the s electrons are held further away from the nucleus.

Consider the electron configurations for the elements in family 7A (the halogens).

Table 9.13: Electron Configuration for Group 7A Elements

Element Electronic Configuration

Fluorine (F) ls 2 2s 2 2p 5

Chlorine (CI) ls 2 2s 2 2p 6 3s 2 3p 5

Bromine (Br) ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5

Iodine (/) ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d w 4p 6 5s 2 4d w 5p 5

The elements in group 7A were placed in the same chemical family by Mendeleev because they all have
similar chemistry, that is, they react with other substances in similar ways and their compounds with
the same metals have similar properties. Now we know why they have similar chemical properties . . .
because chemical properties are controlled by the valence electrons and these elements all have the same
configuration of valence electrons.

The trend for chemical reactivity (speed of reaction) for the non-metal families is the reverse of the situation
in the metal families. The metallic families lose electrons and the largest atoms lose their electrons most
easily so the larger the atom, the faster it reacts. In the case of the non-metals, such as the halogens, the
atoms react by taking on electrons. In these families, the smaller atoms have the largest electron affinity
and therefore take on electrons more readily. Hence, in the non-metals, the smaller atoms will be more
reactive.

Lesson Summary

• Core electrons are the inner electrons that are in filled orbitals and sublevels.

• Core electrons are not responsible for chemical reactivity and do not participate in chemical reactions.

• Valence electrons are the outermost electrons, are responsible for determining the properties, and are
the electrons that participate in chemical reactions.

• Because the members of each group in the main group elements has the same number of valence
electrons, there are similar properties and similar trends in chemical reactions found in the group.

• For metals, chemical reactivity tends to increase with increases in atomic size because the outermost
electrons (or the valence electrons) are further away from the nucleus and therefore have less attraction
for the nucleus.

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Review Questions

1. What is the difference between valence electrons and core electrons?

2. Why would the valence electrons be responsible for the chemical reactivity?

3. Which of the following pairs of elements would have the greatest similarities in terms of chemical
properties?

(a) Ca and Ca 2+

(b) Ca and Mg

(d) Ca and Ar

4. What is the correct number of core electrons in the phosphorous atom?

(a) 3

(b) 5

(d) 15

5. What is the correct number of valence electrons in the iodine atom?

(a) 4

(b) 5

(d) 7

6. For the valence electrons in Group 6A, what conclusions can you draw about the trend in chemical
reactivity?

7. For the valence electrons in Group 7A, what conclusions can you draw about the trend in chemical
activity?

Further Reading / Supplemental Links

Website with lessons, worksheets, and quizzes on various high school chemistry topics.

• Lesson 3-8 is on Lewis Dot Structures.

• Lesson 1-8 is on Elemental Names and Symbols.

• http : //www . f ordhamprep . org/gcurran/sho/sho/lessons/lesson31 . htm

Vocabulary

noble gas core When working with noble gas electronic configurations the core electrons are those
housed in the noble gas symbolic notation.

9.5 Transition Elements

Lesson Objectives

• Define transition metals.

• Explain the relationship between transition metals and the d sublevels.

• State the periods that contain transition metals.

• Write electron configurations for some transition metals.

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Introduction

The main group elements are actually in two groups: the s block and the p block. These two, let's call them
sub groups, are separated in the periodic table by a special group of highly colorful compounds known as
the transition metals. The transition metals represent groups 3 through 12 of the diagram below and are
formed as electrons are filling in the d orbitals. This is why they are referred to as the d block. We will
look at all of this in the lesson that follows. Let's begin!!

Groups (or families)

>rk

14 15

d block

3 4 5 6 7 8 9 10 11 12

■o 3
a

oj 4

3Cl

■

Elements Whose Atoms are Filling d Sublevels

The elements in groups 3 through 12 are called the transition metals. Sometimes this block of elements
are referred to as the d block. They are called d block elements because the electrons being added in this
block of elements are being added to the d orbitals. Look at the electron configurations for Scandium (Sc),
Titanium (Ti), and Vanadium (V), the first three transition metals of the first row in the d block.

215 c:ls 2 2s 2 2p (i 3s 2 3p <i 4s 2 3d 1
22Ti:ls 2 2s 2 2p 6 3s 2 3p 6 As 2 3d 2
23V:ls 2 2s 2 2p s 3s 2 3p s 4s 2 3d 3

Notice that as the atomic number increases, the number of electrons in the d sublevels increases as well.
This is why the phrase d block was started for the transition metals. The transition metals were given
their name because they had a place between the 2A group (now Group 2) and the 3A (now Group 13) in
the main group elements. Therefore, in order to get from calcium to gallium in the periodic table, you had
to transition your way through the first row of the d block ( Sc — » Zn). To get from strontium to indium,
you had to transition your way through the second row of the d block ( Y — » Cd) .

One interesting point about the transition metals that is worth mentioning is that many of the compounds
of these metals are highly colored and used in dyes and paint pigments. Also, metallic ions of transition
elements are responsible for the lovely colors of many gems such as jade, turquoise and amethyst.

Sample question: Write the electron configuration code for Fe.

Solution:

26Fe: 1 s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6

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Transition Metals Occur in Periods 4 — 7

As stated before, transition metals are also called the d block because they fill up the d sublevels. Therefore,
the first three periods do not have transition elements because those energy levels do not have d sub- shells.
The transition metals have atomic numbers greater than 20 because the first d electron appears in element
#21 after the Is, 2s, 2p, 3s, 2>p, and 4s orbitals are filled.

Croups (or families)

14 15

(I block

4 5 6 7 8 9 10 11 12

13 3

<u 4

t OIOCK

Lesson Summary

• Transition metals are those from the d block or those from Groups 3 through 12.

• Compounds of transition metals typically are those that are highly colored.

• Transition metals are found in Periods 4 through 7.

Further Reading / Supplemental Links

• http : //learner . org/resources/series61 . html

Review Questions

1. Write the electron configuration for zirconium, Zr.

2. How many valence electrons does zirconium, Zr, have?

3. Write the noble gas electronic configuration for platinum, Pt.

4. How many valence electrons does platinum, Pt, have?

5. Why do the d block elements only start in the fourth period?

6. What do copper, silver, and gold have in common as far as their electron configuration?

7. Which of these is the electron configuration for nickel?

(a) [Kr]3d 8 4s 2

(b) [Kr]3J 10

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(d) [Ar]3d 10

8. How many d electrons are there in the electronic configuration for ruthenium?

(a)

(b) 6

(d) 17

9. Write the electron configuration for Iridium, Ir.

10. What are the valence electrons for Iridium, Ir?

11. Write the noble gas electron configuration for mercury, Hg.

12. How many valence electrons does mercury, Hg, have?

Vocabulary

transition metal Groups 3 through 12 or the d block of the periodic table.

Labs and Demonstrations for Electron Configuration
Teacher's Pages for Paramagnetism Lab

a————-

After discussing the electron configurations that produce paramagnetism, the student will investigate the
paramagnetism (or lack thereof) with several compounds.

Lab Notes Capsules for this lab can be pre-loaded with the manganese salts. This will save time with a
large class, and is safer. Mn02 can stain the skin and is a mild oxidizing agent. MnO~ A is a strong oxidizing
agent and can cause burns.

If you have trouble with the procedure involving the balance due to balance sensitivity, you may try placing
the gel capsule onto a smooth surface and bring a strong magnet towards it. The capsule may exhibit an
attraction, repulsion, or no reaction. The paramagnetism of the material may be gauged ordinally.

Answers to Pre-Lab Questions

1. Paramagnetism is a weak attraction of a substance for a magnetic field due to the presence of unpaired
electrons. The unpaired electrons each exert a weak magnetic field due to spin, and the spins are
additive when unpaired.

2. Paramagnetism is a weak attraction for a magnetic field, while diamagnetism is a weak repulsion to
a magnetic field due to opposing electron spins within an orbital.

3. The presence of d sublevel in transition metals gives nine opportunities for the sublevel to have an
unpaired electron, not accounting for d 4 or d 9 promotion. This is more than the p sublevel or the s
sublevel.

Lab - Paramagnetic Behavior of Manganese Compounds

Background Information

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When we construct electron configurations for the atoms and ions in the periodic table, one of the cardinal
rules that chemists observe is the pairing of electrons in orbitals whenever possible. Substances with
paired electrons display diamagnetism, a phenomenon that causes that material to be slightly repelled by
a magnetic field. Atoms with unpaired electrons, known as paramagnetic substances, are weakly attracted
to magnetic fields. One interesting example of a molecule that displays paramagnetism is the oxygen
diatomic molecule. While a Lewis dot structure can be constructed pairing all twelve valence electrons
into an oxygen - oxygen double bond with two pairs of non-bonded electrons on each oxygen atom, the
molecule displays paramagnetism when poured between the poles of a strong magnet, attracted to the
magnetic field.

Pre-Lab Questions

1. Explain what is meant by the term paramagnetic.

2. Explain how paramagnetism differs from diamagnetism.

3. Why might paramagnetic transition metal compounds be more prevalent than those of main group
elements?

Purpose

The purpose of this experiment is to demonstrate the effects of a magnetic field on a paramagnetic material.
Apparatus and Materials

• MnC>2

• Mn 2 03
. KMnO^

• Styrofoam cups

• Magnets

• Empty gel capsules

Safety Issues

Manganese oxide is a strong oxidant and can irritate tissues. Potassium permanganate is also a powerful
oxidizing agent and can irritate skin The use of eye protection is strongly recommended.

Procedure

Place about 1.0 gram of each of the materials to be tested in empty gel capsules. The capsules are then
placed on the base of an inverted Styrofoam cup. A second Styrofoam cup is then placed beneath the first
cup. The entire assembly is then placed on a top loading balance. Obtain the mass of the cup and gel
capsule assembly. Now bring the magnet to the cup/capsule assembly and reassess the mass. Note the
decrease in mass as the magnet attracts the capsule contents.

Table 9.14: Data

With Magnet Without Magnet

Mass of the

MnOi /capsule/cup

assembly

Mass of the

Mn-i O3 /capsule/cup

assembly

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Table 9.14: (continued)

With Magnet

Without Magnet

Mass of the

ATMnC^/capsule/cup

assembly

Post-Lab Questions

1 . Which of the samples displayed the largest mass difference?

2. How many unpaired electrons does Manganese possess in MnO^^-

In Mrc 2 03?
In KMnO A l

9.6 Lanthanides and Actinides

Lesson Objectives

• Define the lanthanides and actinides.

• Place the lanthanides and actinides in the periodic table.

• Explain the importance of both the lanthanides and actinides.

• Write electron configurations for lanthanides and actinides.

Introduction

To complete our look at the periodic table there is one more group we have to consider. This group holds its
own unique position in any pictorial representation of the periodic table. The two rows that are generally
placed underneath the main periodic table are called the lanthanide series and the actinide series. These
two rows are produced when electrons are being added to f orbitals. Therefore, this block of elements are
referred to as the "f block". The lanthanides are also occasionally referred to the rare earth elements.
The f block, as shown in the figure below, are the two rows in red.

Croups (or families)

14 15 16

d block

3 4 5 6 7 8 9 10 11 12

■S 3

<u 4

2rf

f block

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Lanthanides and Actinides are Elements Filling the f Sublevels

There is one group that we have neglected to mention throughout this chapter. This group belongs to a
special group found almost disjointed, if you like, from the periodic table. They are given the names the
Lanthanide Series and the Actinide Series or the lanthanides and the actinides. The lanthanides are
an important group of elements. Most of them are formed when uranium and plutonium undergo nuclear
reactions. The elements of the lanthanide series are also known as the rare earth elements. Both the
lanthanides and the actinides make up what are known as the inner transition series. The / block is given
this name because if the f block were placed in its proper numerical position in the periodic table, it would
be in the transition metals between groups 2 and 3.

The lanthanide series includes elements from number 58 to 71, which is 14 elements. The / sub-level
contains seven orbitals and each orbital will hold two electrons. Therefore, it is possible to place 14
electrons in the 4/ sub-level.

The lanthanide series fills the 4/ sublevel as you move from cerium (Ce) to lutetium (L«). The same holds
for the actinide series that runs from atomic number 90 through to number 103, again 14 elements. Thus,
as you move from thorium (Th) at element number 90, you begin to fill up the 5/ sublevel and continue
to fill up the 5/ sublevel until you finish the actinide series at lawrencium (Lr).

14 15 16

HYDROGEN

4A 5A 6A

HELIUM

imam

■nuin

8 9 10

BOHON

CARBON

NITROGEN

OXYGEN

FLUORINE

NEON

MAGNESIUM

SODIUM

1 »B 1

ALUMINUM

SILICON

PHOSPHORUS

SULFUR

CHLORINE

ARGON

POTASSIUM

CALCIUM

SCANDIUM

TITANIUM

VANADIUM

CHROMIUM

MANGANESE

IRON

COBALT

NICKEL

COPPER

ZINC

GALLIUM

GERMANIUM

ARSENIC

SELENIUM

BROMIUM

KRYPTON

A g

RUBIDIUM

STR0N7.UM

YTTRIUM

ZIRCONIUM

NIOBIUM

MOLYBDENUM

TECHNETIUM

RUTHENIUM

RHODIUM

PALLADIUM

SILVER

CADMIUM

INDIUM

TIN

ANTIMONY

TELLURIUM

IODINE

XENON

57-71

La-Lu

H g

CESIUM

BARIUM

LANTHANIDES

HAFNIUM

TANTALUM

TUNGSTEN

RHENIUM

OSMIUM

IRIDIUM

PLATIUM

GOLD

MERCURY

THALUUM

LEAD

BISMUTH

POLONIUM

ASTATINE

RADON

89-103

Ac-Lr

104

105

106

107

108

109

110

111

112

113

Uut

114

Uuq

115

Uup

116

Uuh

117

Uus

118

Uuo

FRANC IUM

RADIUM

FIUTHERFOFIDUM

DUBNIUM

SEABORGIUM

BOHRIUM

HA5SIUM

MEITNERIUM

«..«, .,,».

COPERNICIUM

UNUNTRIUM

mammm

UNUNPENTIUU

i „.,...

UHUNSEPII..

UN.K0C1IUM

LANTHANIDES

LANTHANUM

CERIUM

PHASE0OYM.UM

NEODYMIUM

PROMETHIUM

SAMARIUM

EUROPIUM

GADOLINIUM

TERBIUM

DYSPROSIUM

HOLMIUM

ERBIUM

THULIUM

YTTERBIUM

LUTETIUM

ACTINIDES

100

101

102

103

ACTINIUM

THORIUM

PROTACTINIUM

URANIUM

NEPTUNIUM

PLUTONIUM

AMERICIUM

CURIUM

BERKELIUM

CALIFORNIUM

EINSTEINIUM

FERMIUM

MENOELEVIUM

NOBELUJM

immam

Lanthanides and Actinides Vary in Electron Filling Order

The lanthanides and the actinides make up the f block of the periodic table. The lanthanides are the
elements produced as the 4/ sublevel is filled with electrons and the actinides are formed while filling the
5/ sublevel. Generally speaking, the lanthanides have electron configurations that follow the Aufbau rule.
There are some variations, however, in a few of the lanthanide elements. We will expand a tiny portion
of the periodic table below to show what happens to some of the electron configurations in the lanthanide
and actinide series.

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306

39.098

40.Q7B

44.956

47.867

50.942

POTASSIUM

CALCIUM

SCANDIUM

TITANIUM

VANADIUM

B5.46B

37.62

88.906

91.224

92.905

RUBIDIUM

STRONTIUM

YTTRIUM

ZIRCONIUM

NIOBIUM

57-71

La-Lu

\Hf

CESIUM

137.327
BARIUM

LANTHANIDES

HAFNIUM

TANTALUM

89-103

l 104 \

105

Ac-Lr

223.020

FRANCIUM

226.0254
RADIUM

ACTINIDES

Ve3.ii3 \

RUTHKfORDlim

262.114
DUBNIUM

. LANTHANUM

CERIUM

PRASEODYMIUM

NEODVMIUM

PRQMETHIUM

SAMARIUM

Ell

EUROPIUM

GADOLINIUM

227.027
i ACTINIUM

232.038
THORIUM

231.036
PROTACTINIUM

2 3 B. 029
URANIUM

237.01a
NEPTUNIUM

244.06+
PLUTONIUM

243.061
AMERICIUM

247.070
CURIUM

Look, for example, at the electron configuration for cerium, the first element of the lanthanide series.

Cerium, Ce, is element number 58.

58Ce: ls 2 2i- 2 2p 6 3.s' 2 3p 6 4i 2 3d 10 4p 6 5i 2 4rf 10 5p 6 6i' 2 5rf 1 4/ 1

58Ce:[Xe]6s 2 5d 1 4f 1

Now look at the electronic configuration for praseodymium, an element used in the making of aircraft
engines but also in lighting for making movies. Praseodymium, Pr, is element number 59 and has the
following electron configuration.

59Pr: l.? 2 2.s 2 2p 6 3s 2 3p 6 4i 2 3d 10 4p 6 5.s 2 4rf 10 5p 6 6i 2 4/ 3

59Pr:[Xe]6i 2 4/ 3

Notice the d electron is no longer a part of the electron configuration. There are three lanthanide metals
that have properties similar to the d block. These are cerium, Ce, lutetium, Lu, and gadolinium, Gd. All
of these metals contain a d electron in their electron configuration. The rest, like praseodymium, simply
fill the 4/ sublevel as the atomic number increases.

Unlike the lanthanide family members, most of the actinide series are radioactive. Most of the elements
in the actinide series have the same properties as the d block. Members of the actinide series can lose
multiple numbers of electrons to form a variety of different ions. Table 9.15 shows the noble gas electron
configuration for the elements of the actinide series.

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Table 9.15: Noble Gas Electron Configuration for the Actinide Series

Element

Electron Configuration

Thorium (Th)
Protactinium (Pa)
Uranium (U)
Neptunium (Np)
Plutonium (Pu)
Americium (Am)
Curium (Cm)
Berkelium (Bk)
Californium (Cf)
Einsteinium (Es)
Fermium (Fm)
Mendelevium (Md)
Nobelium (No)
Lawrencium (Lr)

[Rn
[Rn
[Rn
[Rn
[Rn
[Rn
[Rn
[Rn
[Rn
[Rn
[Rn
[Rn
[Rn
[Rn

7s 2 M 2

7s 2 5f 2 6d 1

7s 2 hf<od 1

7s 2 5/ 4 6d 1

7s 2 5f

7s 2 5f 7 6d 1

7s 2 5f

7s 2 5f w

7s 2 5f u

7s 2 5f 12

7s 2 5f 13

7s 2 5f 14

7s 2 5f M 6d 1

Lesson Summary

The lanthanide and actinide series make up the inner transition metals.

The lanthanide series fill up the 4/ sublevel and the actinide series fill up the 5/ sublevel.

The first, middle, and last member of the lanthanide series have properties of the f block and the d

block.

Many of the actinide series have properties of both the d block and the f block elements.

Review Questions

1. Why are the f block elements referred to by some as inner transition elements?

2. What do europium and americium have in common as far as their electron configuration?

3. What is the electron configuration for Berkelium?

(a) [Xe]7s 2 5/ 9

(b) [Xe^s^fM 1

(d) [Rv\7s 2 5fQd l

4. How many f electrons are there in the electron configuration for einsteinium?

(a)

(b) 11

(d) 25

5. Write the electron configuration for Ytterbium, Yb.

6. What are the valence electrons for Ytterbium, Yb? What periods and sublevels are they in?

7. Write the noble gas electronic configuration for uranium, U.

8. What are the valence electrons for uranium, U? What periods and sublevels are they in?

9. Write the electron configurations for neptunium and then for plutonium. Now write an explanation
for what seems to be happening.

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Further Reading / Supplemental Links

• http : //en . wikipedia . org/wiki/Group_number_of _lanthanides_and_act inides

• http : //en . wikipedia . org/wiki/Lanthanide

Vocabulary

lanthanide The rare earth elements found in the first period of the / block. These elements fill up the
4/ sublevel.

actinide The elements found in the second period of the / block. These elements fill up the 5/ sublevel.

Image Sources

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Chapter 10

Trends on the Periodic Table

10.1 Atomic Size

Lesson Objectives

Define atomic radius.

State the boundary issue with atomic size.

Describe measurement methods for atomic size.

Define the shieiding effect.

Describe the factors that determine the trend of atomic size.

Describe the general trend in atomic size for groups and for periods.

Describe the trend of atomic radii in the rows in the periodic table.

Describe how the trend of atomic radii works for transition metals.

Use the general trends to predict the relative sizes of atoms.

Introduction

In the periodic table, there are a number of physical properties that are not really "similar" as it was
previously defined, but are more trend-like. This means is that as you move down a group or across a
period, you will see a trend-like variation in the properties. There are three specific periodic trends that we
will discuss. The first three lessons of this chapter are devoted to the trend in atomic size in the periodic
table. Following this we will discuss ionization energy and electron affinity. Each of these trends can be
understood in terms of the electron configuration of the atoms.

The actual trends that are observed with atomic size have to do with three factors. These factors are:

1. The number of protons in the nucleus (called the nuclear charge).

2. The number of energy levels holding electrons (and the number of electrons in the outer energy level).

3. The number of electrons held between the nucleus and its outermost electrons (called the shielding
effect).

Comparative Atomic Sizes. (Source: CK-12 Foundation. CC-BY-SA)
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Family
1A 2A 3A 4A 5A 6A 7A 8A

Be /R ,C 7 N ,0 y V x Nc

/EKJ ,,- " /<- /i" /" ,1

Li)

g 3

Al ^ Q ^ ^ ©

K Ca Ga Ge As

Sc Bt Kr

5 Rb

Sr In Sri Sb Te 1

6 Cs

Ba Tl Pb Bi Po At Rn

Atoms Have No Definite Boundary

The region in space occupied by the electron cloud of an atom is often thought of as a probability distri-
bution of the electrons and therefore, there is no well-defined "outer edge" of the electron cloud. Atomic
size is defined in several different ways and these different definitions often produce some variations in the
measurement of atomic sizes.

Atomic radius of H2

311

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Because it is so difficult to measure atomic size from the nucleus to the outermost edge of the electron cloud,
chemists use other approaches to get consistent measurements of atomic sizes. One way that chemists define
atomic size is by using the atomic radius. The atomic radius is one-half the distance between the centers
of a homonuclear diatomic molecule (a diatomic molecule means a molecule made of exactly two atoms
and homonuclear means both atoms are the same element). The figure below represents a visualization of
the atomic size definition.

The image on the right is a visual representation of the atomic radius of a hydrogen atom. The measurement
would be taken as one-half the distance between the nuclei of the hydrogen atoms in a diatomic hydrogen
molecule.

How do we measure the size of the atom? Ernest Rutherford is famous for his experiments bombarding gold
foil with alpha particles. The gold foil experiment by Rutherford, first done in 1911, is of particular interest
to us in this unit because it was this experiment that first gave science an approximate measurement for
the size of the atom. He was able, using technology available in the early part of the 1900s, to determine
quantitatively that the nucleus had an approximate size of 4 x 10 -12 cm. The size of the atom is slightly
larger, approximately 2 X 10~ 8 cm in diameter.

Atomic Size in a Column Increases from Top to Bottom

Let's now look at the atomic radii or the size of the atom from the top of a family or group to the bottom.
Take, for example, the Group 1 metals. Each atom in this family (and all other main group families) has
the same number of electrons in the outer energy level as all the other atoms of that family. Each row
(period) in the periodic table represents another added energy level. When we first learned about principal
energy levels, we learned that each new energy level was larger than the one before. Energy level 2 is larger
than energy level 1, energy level 3 is larger than energy level 2, and so on. Therefore, as we move down
the periodic table from period to period, each successive period represents the addition of a larger energy
level. It becomes apparent that as we move downward through a family of elements, that each new atom
has added another energy level and will, therefore, be larger.

Table 10.1:

Element # of protons Electron Configuration # of energy levels

Li 3 [He] 2s 1 2

Na 11 [Ne] 3s 1 3

K 19 [Ar] 4s l 4

Rb 37 [Kr] 5s 1 5

Cs 55 [Xe] 6s 1 6

One other contributing factor to atomic size is something called the shielding effect. The protons in the
nucleus attract the valence electrons in the outer energy level because of opposite electrostatic charges.
The strength of this attraction depends on the size of the charges, the distance between the charges, AND
the number of electrons in-between the nucleus and the valence electrons. The core electrons shield the
valence electrons from the nucleus. The presence of the core electrons weakens the attraction between
the nucleus and the valence electrons. This weakening of the attraction is called the shielding effect. The
amount of shielding depends on the number of electrons between the nucleus and the valence electrons.
The strength with which the nucleus pulls on the valence electrons can pull the valence shell in tighter
when the attraction is strong and not so tight when the attraction is weakened. The more shielding that
occurs, the further the valence shell can spread out.

For example, if you are looking at the element sodium, it has the electron configuration:
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Na ls 2 2s 2 2p 6 3s 1

The outer energy level is n = 3 and there is one valence electron but the attraction between this lone
valence electron and the nucleus that has those 11 protons is shielded by the other 10 inner (or core)
electrons.

When we compare an atom of sodium to one of cesium, we notice that the number of protons increases
as well as the number of energy levels occupied by electrons. There are also many more electrons between
the outer electron and the nucleus, thereby shielding the attraction of the nucleus.

Cs ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 1

The outermost electron, 6s 1 , therefore, is held very loosely. In other words, because of shielding, the
nucleus has less control over this 6s 1 electron than it does over a 3s 1 electron. The result of all of this is
that the atom's size will be larger. Table 2 gives the values for the atomic radii for the group 1 metals plus
a visual representation to appreciate the size change in a group in the periodic table.

• Cs

*Rb

'Na

, Li

Atomic Radii
(in pm)

100-

50-

■ ■

1 1 1 1 1 1

3 11 19 37

Number of Protons in Nucleus

Table 2: Atomic Radii Values for Group 1 Metal (measurement units for atomic radii are picometers
(pm) or 1 X 10~ 12 meters)

Table 10.2:

Element

Atomic radii

Visual

123 pm

313

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Table 10.2: (continued)

Element Atomic radii Visual

Na 157 pm

K 203 pm

Rb 216 pm

Cs 235 pm

What is true for the Group I metals is true for all of the groups, or families, across the periodic table.
As you move downward in the periodic table through a family group, the size of the atoms increases. For
instance, the atoms that are the largest in the halogen family are bromine and iodine (since astatine is
radioactive and only exists for short periods of time, we won't include it in the discussion).

Sample question: Which of the following is larger? Explain.

(a) As or Sb

(b) Ca or Be

(a) Sb because it is below As in Group 15.

(b) Ca because it is below Be in Group 2.

As noted earlier for the main group metals, the outermost energy level in the electron configuration is
indicated by the period number. So the element magnesium (Z = 12), is in period 3, group 2. According
to this, we can say that there are 3 energy levels with 2 electrons in the outermost energy level. Let's look
at the electron configuration for magnesium.

Mg : ls 2 2s 2 2p 6 3s 2

Moving from magnesium to strontium, strontium is in the 5th period of group 2. This means that there
are two electrons in the valence energy level. Strontium also has electrons occupying five energy levels.

Sr: ls 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d l0 4p 6 5s 2

You can imagine that with the increase in the number of energy levels, the size of the atom must increase.
The increase in the number of energy levels in the electron cloud takes up more space.

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Therefore, the trend within a group or family on the periodic table is that the atomic size increases with
increased number of energy levels. The periodic table below shows the trend of atomic size for groups.
The arrow indicates the direction of the increase.

(ft

ca
<u

i—
o

Atomic Size

■

Atomic Size in a Period Decreases from Left to Right

In order to determine the trend for the periods, we need to look at the number of protons (nuclear charge),
the number of energy levels, and the shielding effect. For a row in the periodic table, the atomic number
still increases (as it did for the groups) and thus the number of protons would increase. When we examine
the energy levels for period 2, we find that the outermost energy level does not change as we increase the
number of electrons. In period 2, each additional electron goes into the second energy level. So the number
of energy levels does not go up. As we move from left to right across a period, the number of electrons in
the outer energy level increases but it is the same outer energy level. Table 10.3 shows the electron
configuration for the elements in period 2.

Table 10.3: Electronic Configuration for Elements in Row 2

Element

# of protons

Electron Configuration

Lithium (Li)
Beryllium (Be)
Boron (B)
Carbon (C)
Nitrogen (N)
Oxygen (O)
Fluorine (F)

5
6

Is^s 1

ls 2 2s 2

ls 2 2s 2 2p 1

ls 2 2s 2 2p 2

ls 2 2s 2 2p 3

ls 2 2s 2 2p*

ls 2 2s 2 2p b

Looking at the elements in period 2, the number of protons increases from lithium with three protons, to
fluorine with nine protons. Therefore, the nuclear charge increases across a period. Meanwhile, the number
of energy levels occupied by electrons remains the same. The numbers of electrons in the outermost
energy level increases from left to right along a period but how will this affect the radius? We know
that every one of the elements in row #2 has two electrons in their inner energy level (2 core electrons).
The core electrons shield the outer electrons from the charge of the nucleus. With lithium, there are
two core electrons and one valence electron so those two core electrons will shield the one outer electron.
In beryllium, there are four protons being shielded by the Is 2 electrons. With the increasing number
of protons attracting the outer electrons and the same shielding from the core electrons, the valence

315

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electrons are pulled closer to the nucleus, making the atom smaller.

In group 7A, the first element in the group is fluorine. With fluorine, there are 9 protons and 9 electrons.
The electronic configuration is ls 2 2s 2 2p 5 . However, there are still the same core electrons as with lithium
and beryllium, that is, the Is 2 electrons. Since there are more protons, there is an increase in the nuclear
charge. With an increase in nuclear charge, there is an increase in the pull between the protons and the
outer level, pulling the outer electrons toward the nucleus. The amount of shielding from the nucleus does
not increase because the number of core electrons remains the same (Is 2 for this period). The net result is
that the atomic size decreases going across the row. In Table 10.4, the values are shown for the atomic
radii for the row starting at lithium and ending with fluorine plus a visual representation to appreciate the
size change in a group in the periodic table.

,Li

,Be

Atomic Radii

(in pm)

'c .

► N ,

>° ,

'tie

1 1 1

3453789 10

Number of Protons in Nucleus

Table 10.4: Atomic Radii Values for Period 2

Element

Atomic radii

Visual

123 pm

111 pm

86 pm

77 pm

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316

Table 10.4: (continued)

Element

Atomic radii

Visual

74 pm

73 pm
72 pm

Let's add this new trend to the periodic table. Look at the diagram below of our new "periodic trend
table". In the diagram you will notice that the trend arrow for the period shows the atomic radii increase
going from right to left, which is the same as decreasing from left to right.

Considering all the information about atomic size, you will recognize that the largest atom on the periodic
table is all the way to the left and all the way to the bottom, francium, #87, and the smallest atom is all
the way to the right and all the way to the top, helium, #2.

The fact that the atoms get larger as you move downward in a family is probably exactly what you expected
before you even read this section, but the fact that the atoms get smaller as you move to the right across
a period is most likely a big surprise. Make sure you understand this trend and the reasons for it.

For the Transition Elements, the Trend is Less Systematic

A general trend for atomic radii in the periodic would look similar to that found in the diagram below.
The elements with the smallest atomic radii are to the upper right; those with the largest atomic radii are
to the lower left.

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(A
<D
(A
(d
CD
•—
o
c

Increases

Atomic Size

Until now, we have worked solely with the main group metals. Let's look at our three factors and see how
these factors fit the transition metal series. Looking at the first row of the transition metals, the 3d row,
Table 10.5 shows the number of protons in each of the 10 elements in this row. The number of protons is
increasing so the nuclear charge is increasing.

Table 10.5: Summary of Data for 3d Metals

Element

# of protons

Electron Configuration

Scandium (Sc)
Titanium (77)
Vanadium (V)
Chromium (Cr)
Manganese (Mn)
Iron (Fe)
Cobalt (Co)
Nickel (Ni)
Copper (Cm)
Zinc (Zn)

21
22
23
24
25
26
27
28
29
30

[Ar

3d l 4s 2

[Ar

3d 2 4s 2

[Ar

3d 3 4s 2

[Ar

3d b 4s 1

[Ar

3d 5 4s 2

[Ar

3d 6 4s 2

[Ar

3d 7 4s 2

[Ar

3d 8 4s 2

[Ar

3d w 4s 1

[Ar

3d w 4s 2

The number of electrons are increasing, but in a particular way. We know that as the number of electrons
increases going across a period, there is more pull of these electrons toward the nucleus. However, with
the ^-electrons, there is some added electron-electron repulsion. Take a look at Table 10.5 and note the
unusual electron configuration of chromium.

In chromium, there is a promotion of one of the 4s electrons to half fill the 3d sublevel, the electron-electron
repulsions are less and the atomic size is smaller. The opposite holds true for the latter part of the row.
Table 10.6 shows the first row of the transition metals along with their size.

Table 10.6: Atomic Radii for 3d Metals

Element

Atomic radii (pm)

Scandium (Sc)
Titanium (77)
Vanadium (V)
Chromium (Cr)
Manganese (Mn)

164
147
135
129
137

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318

Table 10.6: (continued)

Element

Atomic radii (pm)

Iron (Fe)

Cobalt (Co)
Nickel (Ni)
Copper (Cu)
Zinc (Zn)

126
125
125
128
137

The graph of the number of protons versus the atomic radii is shown below.

200

150

atomic radii (pm)

100

H 1 1 1 h

-A

H 1 1 1 1 1 1 1 1 1 1 1 h

25 3i

Number of protons

Graphing the atomic number (or the number of protons) versus the atomic radii, we can see the trend in
the 3d transition metals isn't quite as systematic as with the main group elements.

Lesson Summary

• Atomic size is the distance from the nucleus to the valence shell where the valence electrons are
located.

• The electrons surrounding the nucleus exist in an electron cloud.

• You can predict the probability of where the electron is but not its exact location.

• Atomic size is difficult to measure because it has no definite boundary.

• Atomic radius is a more definite and measureable way of defining atomic size. It is the distance from
the center of one atom to the center of another atom in a homonuclear diatomic molecule.

• Rutherford led the way to determining the size of the atom with his gold foil experiment.

• Mass spectrometers and other machines are available to measure masses and determine structure
directly.

• Atomic size is determined indirectly.

• There are three factors that help in the prediction of the trends in the periodic table: number of
protons in the nucleus, number of shells, and shielding effect.

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• The atomic size increases from the top to the bottom in any group as a result of increases in all of
the three factors. *As the number of energy levels increases, the size must increase.

• Going across a period (from left to right), the number of protons increases and therefore the nuclear
charge increases. *Going across a period, the number of electron energy levels remains the same but
the number of electrons increases within these energy levels. Therefore the electrons are pulled in
closer to the nucleus.

• Shielding is relatively constant since the core electrons remain the same.

• The trend in the periodic table is that as you move across the periodic table from left to right, the
atomic radii decrease. This trend is not as systematic for the transition metals because other factors
come into play.

Review Questions

1. Why is the atomic size considered to have "no definite boundary"?

2. How is atomic size measured?

(a) using a spectrophotomer

(b) using a tiny ruler (called a nano ruler)

(d) directly

3. Draw a visual representation of the atomic radii of an iodine molecule.

4. Which of the following would be smaller?

(a) In or Ga

(b) K or Cs

5. Explain in your own words why Iodine is larger than Bromine.

6. What three factors determine the trend of atomic size going down a group?

7. What groups tend to show this trend?

8. Which of the following would have the largest atomic radii?

(a) Si

(b) C

(d) Pb

9. Which of the following would have the smallest atomic radius?

(a) 2s 2

(b) 4s 2 4p 3

(d) As 2

10. Arrange the following in order of increasing atomic radii: Tl,B,Ga,Al,In.

11. Arrange the following in order of increasing atomic radii: Ge,Sn,C,

12. Which of the following would be larger?

(a) Rb or Sn

(b) Ca or As

13. Place the following in order of increasing atomic radii: Mg,Cl,S,Na.

14. Describe the atomic size trend for the rows in the periodic table.

15. Draw a visual representation of the periodic table describing the trend of atomic size.

16. Which of the following would have the largest atomic radii?

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(a)

(b)

(c)

(d)

M "hi

>\ ill

(a)

en C

(b)

(c)

(d)

17. Which of the following would have the smallest atomic radii?

18. Place the following elements in order of increasing atomic radii: In, Ca, Mg, Sb, Xe.

19. Place the following elements in order of decreasing atomic radii: Al,Ge,Sr,Bi,Cs.

20. Knowing the trend for the rows, what would you predict to be the effect on the atomic radius if an
atom were to gain an electron? Use an example in your explanation.

21. Knowing the trend for the rows, what would you predict to be the effect on the atomic radius if the
atom were to lose an electron? Use an example in your explanation.

Vocabulary

atomic size Atomic size is the distance from the nucleus to the valence shell where the valence electrons
are located.

atomic radius One-half the distance between the centers of the two atoms of a homonuclear molecule.

nuclear charge The number of protons in the nucleus.

shielding effect The core electrons in an atom interfere with the attraction of the nucleus for the out-
ermost electrons.

electron-electron repulsion The separation that occurs because electrons have the same charge.

10.2 Ionization Energy

Lesson Objectives

Define ionization energy.

Describe the trend that exists in the periodic table for ionization energy.

Describe the ionic size trend that exists when elements lose one electron.

Introduction

When we study the trends in the periodic table, we cannot stop at just atomic size. In this section of the
chapter, we will begin an understanding of an important concept, namely ionization energy and recognize
its trend on the periodic table.

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Ionization Energy is the Energy Required to Remove an Electron

Lithium has an electron configuration of ls 2 2s 1 . Lithium has one electron in its outermost energy level.
In order to remove this electron, energy must be added to the system. Look at the equation below:

energy + L/ (g)
ls 2 2s l

Li (g) + e
Is 2

With the addition of energy, a lithium ion can be formed from the lithium atom by losing one electron.
This energy is known as the ionization energy. The ionization energy is the energy required to remove
the most loosely held electron from a gaseous atom or ion. "In the gaseous phase" is specified because in
liquid or solid, other energies get involved. The general equation for the ionization energy is as follows.

energy + A fe)

A Z) +e ~

The higher the value of the ionization energy, the harder it is to remove that electron. We can see a trend
when we look at the ionization energies for the elements in period 2. Table 10.7 summarizes the electron
configuration and the ionization energies for the elements in the second period.

Table 10.7: First Ionization Energies

Element

Electron configuration

First Ionization Energy, IE\

Lithium (Li)
Beryllium (Be)
Boron (B)
Carbon (C)
Nitrogen (N)
Oxygen (O)
Fluorine (F)

[He]

2s'

[He]

2s 2

[He]

2s 2 2p l

[He]

2s 2 2p 2

[He]

2s 2 2p 3

[He]

2s 2 2p A

[He]

2s 2 2p 5

520 kJ/mol
899 kJ/mol
801 kJ/mol
1086 kJ/mol
1400 kJ/mol
1314 kJ/mol
1680 kJ/mol

When we look closely at the data presented in Table 10.7, we can see that as we move across the period
from left to right, in general, the ionization energy increases. At the beginning of the period, with the
alkali metals and the alkaline earth metals, losing one or two electrons allows these atoms to become ions.

energy + Li(g) -> Li + (g) + e~
energy + Mg(g) -» Mg 2+ + 2e~

[He] 2s 1 -> [He] + e~
[Ne] 3s 2 -> [Ne] + 2e~

As we move across the period, the atoms become smaller which causes the nucleus to have greater attraction
for the valence electrons. Therefore, the electrons are more difficult to remove.

A similar trend can be seen for the elements within a family. Table 10.8 shows the electron configuration
and the first ionization energies (IE\) for some of the elements in the first group, the alkali metals.

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Table 10.8: Ionization Energies for Some Group 1 Elements

Element

Electron configuration

First Ionization Energy, IE\

Lithium (Li)
Sodium (No)
Potassium (K)

[He] 2s 1
[He] 3s 1
[He] 4s 1

520 kJ/mol
495.5 kJ/mol
418.7 kJ/mol

Comparing the electron configurations of lithium to potassium, we know that the electron to be removed
is further away from the nucleus. We know this because the n value is larger meaning the energy level
where the valence electron is held is larger. Therefore it is easier to remove the most loosely held electron
because the atom is larger with a greater shielding effect which means that the nucleus has less control
over potassium's outer electron, 4s 1 .

Therefore IE\ for potassium (418.7 kJ/mol) is less than IE\ for lithium (520 kJ/mol).

If a second electron is to be removed from an atom, the general equations are the following:

/(g) + energy -> / 1+ + e
J+ (g) + ener gy ~> j2+ (g) + e ~

IE l
IE 2

Since there is an imbalance of positive and negative charges when a second electron is being removed,
the energy required for the second ionization (IE2) will be greater than the energy required for the first
ionization (IE\). Simply put, IE\ < IE2 < IE3 < IE4.

The Charge on the Nucleus Increases and Size Decreases

So if we look at the ionization energy trend in the periodic table and add it to the trend that exists with
atomic size we can show the following on our periodic table chart.

Ionization Energy Increases

Atomic Size Increases

But why does the ionization energy increase going across a period and decrease going down a group? It has
to do with two factors. One factor is that the atomic size decreases. The second factor is that the effective
nuclear charge increases. The effective nuclear charge is the charge experienced by a specific electron
within an atom. Remember, the nuclear charge was used to describe why the atomic size decreased going
across a period. When we look at the data in Table 10.7 again, we can see how the effective nuclear charge
increases going across a period. Table 10.9 shows the effective nuclear charge along with the ionization
energy for the elements in period 2.

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Table 10.9: Effective Nuclear Charge for Period 2 Main Group Elements

Element

Electron con-
figuration

# of protons # of core elec- Effective nu- Ionization En-
trons clear charge ergy

Lithium (Li)

[He]

2s 1

Beryllium (Be)

[He]

2s 2

Boron (B)

[He]

2s 2 2p 1

Carbon (C)

[He]

2s 2 2p 2

Nitrogen (N)

[He]

2s 2 2p 3

Oxygen (0)

[He]

2s 2 2p 4

Fluorine (F)

[He]

2s 2 2p 5

2
2
2
2
2
2
2

1
2
3

4
5
6
7

520 kJ/mol
899 kJ/mol
801 kJ/mol
1086 kJ/mol
1400 kJ/mol
1314 kJ/mol
1680 kJ/mol

The electrons that are shielding the nuclear charge are the core electrons, or in the case of period 2, the Is 2
electrons. The effective nuclear charge is the difference between the total charge in the nucleus (the number
of protons) and the number of shielded electrons. Notice how as the effective nuclear charge increases, so
does the ionization energy. Overall the general trend for ionization energy is shown below.

Ionization Energy

Inert

;ase<

a -
<u

i— i -

l^n

Sample question:

What would be the effective nuclear charge for CI? Would you predict the ionization energy to be higher
or lower than fluorine?

Solution:

Chlorine has the electronic configuration of:

CI : [Ne] 3s 2 3p 5

The effective nuclear charge is 7, the same as fluorine. Predicting the ionization energy would be difficult
with this alone. The atomic size, however, is larger for chlorine than for fluorine because now there are
three energy levels (chlorine is in period 3). Now we can say that the ionization energy should be lower
than that of fluorine because the electron would be easier to pull off the level further away from the nucleus.
(Indeed, the value for chlorine is 1251 kJ/mol).

Some Anomalies With the Trend in Ionization Energy

There are a few anomalies that exist with respect to the ionization energy trends. Going across a period
there are two ways in which the ionization energy may be affected by the electron configuration. When we
look at period 3, we can see that there is an anomaly observed as we move from the 3s sublevels to the 3p

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324

sublevel. The table below shows the electron configurations for the main group elements in period 3 along
with the first ionization energy for these elements.

Table 4: Ionization Energies for Period 3 Main Group Elements

Element

Electron Configuration

Ionization Energy

Sodium (Na)

1s 2 2s 2 2p e 3s 1

495.9 kJ/mol

Magnesium (Mg}

1s 2 2s 2 2p 6 3s 2

738.1 kJ/mol 1

Aluminum (Al)

1s 2 2s 2 2p 6 3s 2 3p 1

577.9 kJ/mol J

Silicon [Si)

1s 2 2s 2 2p 6 3s 2 3p 2

786.3 kJ/mol

Phosphorus [P)

1s 2 2s 2 2p 6 3s 2 3p 3

1012 kJ/mol

Sulfur [S)

1s 2 2s 2 2p 6 3s 2 3p 4

999.5 kJ/mol

Chlorine [CI)

1s 2 2s 2 2p 6 3s 2 3p 5

1251 kJ/mol

Argon (Ar)

1s 2 2s 2 2p 6 3s 2 3p 6

1520 kJ/mol

f| 3p

In the table we see that when we compare magnesium to aluminum the first IE decreases instead of
increasing as we would have expected. So why would this be so? Magnesium has its outermost electrons
in the s sublevel. The aluminum atom has its outermost electron in the 2>p sublevel. Since p electrons
have just slightly more energy than s electrons, it takes a little less energy to remove that electron from
aluminum. One other slight factor is that the electrons in 3s 2 shield the electron in 3p . These two factors
allow the IE\ for Al to be less than IE\ for Mg.

When we look again at Table 1 below, we can see that the ionization energy for nitrogen seems out of
place.

Table 1 : Ionization Energies for Period 2 Main Group Elements

Element

Electron Configuration

Ionization Energy

Lithium [Li)

[He]2s 1

520 kJ/mol

Beryllium [Be)

[He] 2s 2

899 kJ/mol

Boron [B)

[He] 2s 2 2p 1

801 kJ/mol

Carbon [C)

[He] 2s 2 2p 2

1086 kJ/mol

Nitrogen (N)

[He]2s 2 2p 3

1400 kJ/mol 4

{

Oxygen (0)

[He] 2s 2 2p 4

_
1314kJ/mol

Fluorine (F)

[He]2s 2 2p"

1680kJ/mol

Neon (Ne)

[He] 2s 2 2p 6

2081 kJ/mol

_t_ _L_t_

\\ 2P

HJ_i

While nitrogen has one electron occupying each of the three p orbitals in the 2 nd sublevel, oxygen has

325

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two orbitals occupied by only one electron but one orbital containing a pair of electrons. The greater
electron-electron repulsion experienced by these 2p electrons allows for less energy to be needed to remove
one of these. Therefore, IE\ for oxygen is less than nitrogen.

Lesson Summary

• Ionization energy is the energy required to remove the most loosely held electron from a gaseous
atom or ion. Ionization energy generally increases across a period and decreases down a group. The
effective nuclear charge is the charge of the nucleus felt by the valence electron.

• The effective nuclear charge and the atomic size help explain the trend of ionization energy. Going
down a group the atomic size gets larger and the electrons can be more readily removed, therefore,
ionization energy decreases. Going across a period the effective nuclear charge increase so the electrons
are harder to remove and the ionization energy increases. Once one electron has been removed, a
second electron can be removed but IE\ < IE<i- If a third electron is removed, IE\ < IE2 < IE3 and
so on.

Review Questions

1. Define ionization energy and show an example ionization equation.

2. Draw a visual representation of the periodic table describing the trend of ionization energy.

3. Which of the following would have the largest ionization energy?

(a) Na

(b) Al

(d) He

4. Which of the following would have the smallest ionization energy?

(a) K

(b) P

(d) Ca

5. Place the following elements in order of increasing ionization energy: Na,0,Ca,Ne,K.

6. Place the following elements in order of decreasing ionization energy: N,Si,S,Mg,He.

7. Using experimental data, the first ionization energy for an element was found to be 600 kJ/mol. The
second ionization energy for the ion formed was found to be 1800 kJ/mol. The third ionization energy
for the ion formed was found to be 2700 kJ/mol. The fourth ionization energy for the ion formed
was found to be 11600 kJ/mol. And finally the fifth ionization energy was found to be 15000 kJ/mol.
Write the reactions for the data represented in this question. Which group does this element belong?
Explain.

8. Using electron configurations and your understanding of ionization energy, which would you predict
would have higher second ionization energy: Na or Mg?

9. Comparing the first ionization energy {IE\) of calcium, Ca, and magnesium, Mg, :

(a) Ca has a higher IE\ because its radius is smaller.

(b) Mg has a higher IE\ because its radius is smaller.

(d) Mg has a higher IE\ because its outer sub-shell is full.

(e) they have the same IE\ because they have the same number of valence electrons.

10. Comparing the first ionization energy {IE\) of beryllium, Be, and boron, B, :
www.ckl2.org 326

(a) Be has a higher IE\ because its radius is smaller.

(b) B has a higher IE\ because its radius is smaller.

(d) B has a higher IE\ because its s sub-shell is full.

(e) They have the same IE\ because B has only one more electron than Be.

Further Reading / Supplemental Links

• http://en.wikipedia.org

Vocabulary

ionization energy The energy required to remove an electron from a gaseous atom or ion:

energy + J(g) — > J + (g) + e~ (first ionization energy).

effective nuclear charge The charge on the atom or ion felt by the outermost electrons (valence elec-
trons).

10.3 Electron Affinity

Lesson Objectives

• Define electron affinity.

• Describe the trend for electron affinity on the periodic table.

Introduction

The final periodic trend for our discussion is electron affinity. We have talked about atomic structure,
electronic configurations, size of the atoms and ionization energy. And now, the final periodic trend we
will study is how an atom can gain an electron and the trends that exist in the periodic table.

The Energy Process When an Electron is Added to an Atom

Atoms can gain or lose electrons. When an atom gains an electron, energy is given off and is known as
the electron affinity. Electron affinity is defined as the energy released when an electron is added to a
gaseous atom or ion.

T(g)+e- ^ T- [g)

When most reactions occur that involve the addition of an electron to a gaseous atom, potential energy is
released.

Br {g) + e~ -> Br l ~( g) EA = -325 kJ/mol

[Ar] 4s 2 4p 5 [Ar] 4s 2 4p 6

Let's look at the electron configurations of a few of the elements and the trend that develops within groups
and periods. Take a look at Table 10.10, the electron affinity for the Halogen family.

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Table 10.10: Electron Affinities for Group 7A

Element

Electron configuration

Electron affinity, kJ/mol

Fluorine (F)
Chlorine (CI)
Bromine (Br)
Iodine (7)

[He] 2s 2 2p 5
[Ne] 3s 2 3p 5
[Ar] 4s 2 4p 5
[Kr] 5s 2 5p 5

-328
-349
-325
-295

As you can see, the electron affinity generally decreases (becomes less negative) going down a group because
of the increase in size of the atoms. Remember that the atoms located within a family but lower on the
periodic table are larger since there are more electrons filling more energy levels. For example, an atom of
chlorine is smaller than iodine; or, an atom of oxygen is smaller than sulfur. When an electron is added to
a large atom, less energy is released because the electron cannot move as close to the nucleus as it can in
a smaller atom. Therefore, as the atoms in a family get larger, the electron affinity gets smaller.

There is an exception to this when it involves certain small atoms. Electron affinity for fluorine is less than
chlorine most likely due to the electron-electron repulsions that occur between the electrons where n = 2.
This phenomenon is observed in other families as well. For instance, the electron affinity for oxygen is
less than the electron affinity for sulfur. Electron affinity of all of the elements in the second period is less
than the ones below them due to the fact that the elements in the second period have such small electron
clouds that electron repulsion is greater than that of the rest of the family.

Nonmetals Tend to Have the Highest Electron Affinity

Overall, the periodic table shows the general trend similar to the one below.

Table 5.1: Electron Affinities for Period 4 Main Group Elements

Element

Electron Configuration

Electron Affinity

Potassium (K)

[Ar] 4s 1

-48 kJ/mol

Calcium (Ca)

[Ar]4s 2

-2.4 kJ/mol H__^

Gallium (Ga)

[Ar] 4s 2 4p 1

-29 kJ/mol

Germanium (Ge)

[Ar]4s 2 4p 2

-118 kJ/mol

Arsenic (As)

[Ar]4s 2 4p 3

-77 kJ/mol E_

Selenium (Se)

[Ar]4s 2 4p 4

-195 kJ/mol

Bromine (Br)

[Ar]4s 2 4p 5

-325 kJ/mol

Krypton [Kr)

[Ar] 4s 2 4p 6

kJ/mol

LJ_i

The general trend in the electron affinity for atoms is almost the same as the trend for ionization energy.
This is because both electron affinity and ionization energy are highly related to atomic size. Large atoms
have low ionization energy and low electron affinity. Therefore, they tend to lose electrons and do not
tend to gain electrons. Small atoms, in general, are the opposite. Since they are small, they have high
ionization energies and high electron affinities. Therefore, the small atoms tend to gain electrons and tend
not to lose electrons. The major exception to this rule are the noble gases. They are small atoms and

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328

do follow the general trend for ionization energies. The noble gases, however, do not follow the general
trend for electron affinities. Even though the noble gases are small atoms, their outer energy levels are
completely filled with electrons and therefore, an added electron cannot enter their outer most energy level.
Any electrons added to a noble gas would have to be the first electron in a new (larger) energy level. This
causes the noble gases to have essentially zero electron affinity. This concept is discussed more thoroughly
in the next chapter.

When atoms become ions, the process involves either the energy released through electron affinity or
energy being absorbed with ionization energy. Therefore, the atoms that require a large amount of energy
to release an electron will most likely be the atoms that give off the most energy while accepting an
electron. In other words, non-metals will most easily gain electrons since they have large electron affinities
and large ionization energies; and, metals will lose electrons since they have the low ionization energies
and low electron affinities.

Now let's add this last periodic trend to our periodic table representation and our periodic trends are
complete.

Atomic Size Increases

Ionization Energy Increases

* *

Electron Affinity 1

icreases

IS,

0)
is,

to
b

LU
o

£Z

r s

1 ai

cu
is,
ra
a>

ez
%

T3
a)

—

Note: Both the trend in atomic size and
the trend in ionization energy extend all
the way left to right and top to bottom in
the periodic table, but the trend in electron
affinity only extends through family 7A, the
halogens. The noble gases, family 8A,
essentially have zero electron affinity.

The web site below contains some textual material discussing chemical bonds and also allows you to access
tables of graphs showing the ionization energies and electron affinities of various elements. Chemical Bond
Data (http : //hyperphysics . phy-astr . gsu . edu/hbase/chemical/bondd . html)

The development and arrangement of the periodic table of elements is examined. Video on Demand - The
World of Chemistry - The Periodic Table (http: //www. learner. org/vod/vod_window.html?pid=799)

Lesson Summary

• Electron affinity is the energy required (or released) when an electron is added to a gaseous atom or
ion. Electron affinity generally increases going up a group and increases left to right across a period.

• Non-metals tend to have the highest electron affinities.

Review Questions

1. Define electron affinity and show an example equation.

2. Choose the element in each pair that has the lower electron affinity:

(a) Li or N

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(b) Na or CI

(d) Mg or F

3. Why is the electron affinity for calcium much higher than that of potassium?

4. Draw a visual representation of the periodic table describing the trend of electron affinity.

5. Which of the following would have the largest electron affinity?

(a) Se

(b) F

(d) Br

6. Which of the following would have the smallest electron affinity?

(a) Na

(b) Ne

(d) /?£>

7. Place the following elements in order of increasing electron affinity: Te,Br,S,K,Ar.

8. Place the following elements in order of decreasing electron affinity: S,Sn,Pb,F,Cs.

9. Describe the trend that would occur for electron affinities for elements in period 3. Are there any
anomalies? Explain.

10. Comparing the electron affinity (EA) of sulfur, S , and phosphorus, P, :

(a) S has a higher EA because its radius is smaller.

(b) P has a higher EA because its radius is smaller.

(d) P has a higher EA because its p sub-shell is half full.

(e) they have the same EA because they are next to each other in the periodic table.

Vocabulary

electron affinity The energy required to add an electron to a gaseous atom or ion.

T(g) + e- -» T-(g)

Image Sources

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Chapter 11
Covalent Bonding

11.1 The Covalent Bond

Lesson Objectives

• The student will describe the basic nature of covalent bond formation.

• The student will explain the difference between ionic and covalent bonding.

• The student will state the relationship between molecular stability and bond strength.

Introduction

In ionic bonding, electrons leave metallic atoms and enter non-metallic atoms. This complete transfer
of electrons changes both of the atoms into ions. Often, however, two atoms combine in a way that no
complete transfer of electrons occurs. Instead, electrons are held in overlapping orbitals of the two atoms,
so that the atoms are sharing the electrons. The shared electrons occupy the valence orbitals of both atoms
at the same time. The nuclei of both atoms are attracted to this shared pair of electrons and the atoms
are held together by this attractive force. The attractive force produced by sharing electrons is called a
covalent bond.

Sharing Electrons

In covalent bonding, the atoms acquire a stable octet of electrons by sharing electrons. The covalent
bonding process produces molecular substances as opposed to the lattice structures of ionic bonding. The
covalent bond, in general, is much stronger than ionic bonds and there are far more covalently bonded
substances than ionic substances.

The diatomic hydrogen molecule, H2, is one of the many molecules that are covalently bonded. Each
hydrogen atom has a Is electron cloud containing one electron. These Is electron clouds overlap and
produce a common volume which the two electrons occupy (Figure 11.1).

The shared pair of electrons spend more time between the two atoms than they do in other parts of the
two Is orbitals but the shared pair of electrons occupies all of the Is orbitals of both atoms.

In the simulated probability pattern for the overlapped Is orbitals in an H2 molecule, we see the electrons
are still more likely to be found close to the nucleus than far away, but we also see that they spend more
time between the two nuclei than they do on the far sides of the atoms. The extra time spent between the

331 www.ckl2.org

Figure 11.1: The orbitals of two hydrogen atoms overlap and share the two valence electrons.

two nuclei is the source of the attraction that holds the atoms together in a covalent bond (Figure 11.2).

Figure 11.2: Simulated probability pattern for the overlapped orbitals in .

The diatomic fluorine molecule is also a covalently bonded atom. In the case of fluorine atoms, the atoms
have filled Is orbitals, filled 2s orbitals, and two of the three 2p orbitals are full. Each atom has a half-filled
2p orbital that is available to be overlapped.

Figure 11.3: Showing the orbitals of fluorine with two orbitals full and one orbital half- full and then showing
the half-filled orbital of each atom overlapping.

Figure 11.3 shows the available (half-filled) 2p orbitals of two fluorine atoms overlapping to form a covalent
bond. We use several methods to represent a covalent bond so that we don't have to spend all our time
drawing orbitals. We can represent the bond in the F2 molecule with an electron dot formula (Figure
11.4).

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332

:f:f: :f-F

• •

Figure 11.4: We can also show a covalent bond between atoms with an electron dot formula where the
shared pair of electrons are the bonding electrons or with the bond represented by a dash.

Bond Strength

When atoms that attract each other move closer together, the potential energy of the system (the two
atoms) decreases. When a covalent bond is formed, the atoms move so close together that the atoms
overlap their electron clouds. As the atoms continue to move closer yet, the potential energy of the
system continues to decrease ... to a point. If you continue to move atoms closer and closer together,
eventually the two nuclei will begin to repel each other. If you push the nuclei closer together at this
point, the repulsion causes the potential energy to increase. Each pair of covalently bonding atoms will
have a distance between their nuclei that is the lowest potential energy distance. This position has the
atoms close enough that the attraction between the nucleus of each one and the electrons of the other is
maximum but the nuclei have not begun to repel each other strongly. For bonding atoms, this distance
occurs somewhere after the electron clouds have overlapped. At this position, the atoms are at lowest
potential energy. If the atoms are pushed closer, the potential energy goes up because you are crowding
the nuclei together and if the atoms are pulled apart, potential energy goes up because you are separating
particles that attract each other. Since this is the lowest potential energy position, the atoms will remain
at the distance, bonded together. This distance is called the bond length. The more potential energy that
was given up as this bond formed, the stronger the bond will be. If you want to break this bond, you must
input all the potential energy that was given up as the bond was formed.

The strength of a diatomic covalent bond can be expressed by the amount of energy necessary to break the
bond and produce separate atoms. The energy needed to break a covalent bond is called bond energy
and is measured in kilojoules per mole. Bond energy is used as a measure of bond strength. The bond
strength of HBr is 365 kilojoules per mole. i.e. It would take 365 kilojoules to break all the chemical bonds
in 6.02 x 10 23 molecules of HBr and produce separate hydrogen and bromine atoms. The bond strength of
HCl is 431 kilojoules per mole. Consequently, the bond in HCl is stronger than the bond in HBr.

Molecular Stability

Compounds with high bond strengths are difficult to break up and therefore are stable compounds. When
stable compounds are formed, large amounts of energy are given off so these molecules are in a relatively
low energy state. In order to break the molecule apart, all the energy that was given off must be put back
in. Low energy state bonds are stable and have high bond strength.

Molecules with high bond energy have weak bonds. They did not release much energy when they formed
and so not much energy is needed to the break the molecules back apart. High bond energy means low
bond strength.

Atoms of carbon, hydrogen, and oxygen can be chemically bonded in more than one way. In the molecule
shown below, these atoms are bonded in a way that produces a molecule of glucose.

333 www.ckl2.org

CHO

H-C-OH

I
HO-C-H

I
H-C-OH

CHiOH

+ 6 2 — *►" 6 COi + 6 H 2 + ENERGY

The molecule of glucose can be reacted with six oxygen atoms to produce six molecules of carbon dioxide
and six molecules of water. During the reaction, the atoms of the glucose molecule are rearranged into the
structure of carbon dioxide and water molecules. The bonds in the glucose are broken and new bonds are
formed. As this occurs, potential energy is released because the new bonds have lower potential energy
that the original bonds. The bonds in the products are lower energy bonds and therefore, the product
molecules are more stable.

Some Compounds Have Both Covalent and Ionic Bonds

If you recall the introduction of polyatomic ions, you will remember that the bonds that hold the polyatomic
ions together are covalent bonds. Once the polyatomic ion is constructed with covalent bonds, it reacts
with other substances as an ion. The bond between a polyatomic ion and another ion will be ionic. An
example of this type of situation is in the compound sodium nitrate. Sodium nitrate is composed of a
sodium ion and a nitrate ion. The nitrate ion is held together by covalent bonds and the nitrate ion is
attached to the sodium ion by an ionic bond (Figure 11.5).

covalent

NaN0 3

covalent
ionic

Figure 11.5: The bonding in sodium nitrate, .

Lesson Summary

• Covalent bonds are formed by electrons being shared between two atoms.

• Half-filled orbitals of two atoms are overlapped and the valence electrons shared by the atoms.
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• Bond energy is the amount of energy necessary to break the covalent bond.

• The strength of a covalent bond is measured by the bond energy.

• Stable compounds have high bond energy and unstable compounds have low bond energy.

Review Questions

1. Describe the characteristics of two atoms that would be expected to form an ionic bond.

2. Describe the characteristics of two atoms that would be expected to form a covalent bond.

3. If an atom had a very high bond energy, would you expect it to be stable or unstable?

4. When gaseous potassium ions and gaseous fluoride ions join together to form a crystal lattice, the
amount of energy released is 821 kJ/mol. When gaseous potassium ions and gaseous chloride ions
join together to form a crystal lattice, the amount of energy released is 715 kJ/mol. Which is the
stronger bond, KF or KCll It these two compounds were increasingly heated, which compound would
break apart at the lower temperature?

Further Reading / Supplemental Links

Website with lessons, worksheets, and quizzes on various high school chemistry topics.

• Lesson 4-3 is on Ionic vs. Covalent Bonds.

• http : //www . f ordhamprep . org/gcurran/sho/sho/lessons/lesson43 . htm

Vocabulary

covalent bond A type of chemical bond where two atoms are connected to each other by the sharing of
two or more electrons in overlapped orbitals.

covalent bond strength The strength of a covalent bond is measured by the amount of energy required
to break the bond.

11.2 Atoms that Form Covalent Bonds

Lesson Objectives

• The student identify pairs of atoms that will form covalent bonds.

• The student will draw Lewis structures for simple covalent molecules.

• The student will identify sigma and pi bonds in a Lewis structure.

Introduction

A great deal of importance seems to have been attached to the electron configurations of noble gases or to
the octet of electrons. These structures have been made in some way to seem to be a "desirable" thing for
an atom to have. It is hoped that you do not see atoms and electrons as behaving the way they do because

335 www.ckl2.org

of "wants" and "desires". The bonding of atoms is directly by the laws of nature relating to the tendency
toward minimum potential energy, electrical attraction and repulsion, and the arrangement of electrons
in atoms. As it happens, these laws of nature and energy conditions do favor (in most cases) an octet of
electrons for atoms. In ionic bonding, the atoms acquired this octet by gaining or losing electrons and,
in covalent bonding, as you have seen, the atoms acquire the noble gas electron configuration by sharing
electrons.

Non-Metals Bond with Non- Metals to Form Covalent Bonds

In the discussion of ionic bonds, it was clear that ionic bonds form between metals and non-metals because
the high electron affinity non-metals are able to take electrons away from metals. Metals lose their electrons
readily and have no attraction to add electrons. Since covalent bonds require that electrons be shared,
it becomes apparent that metals will form few if any covalent bonds. Metals simply do not hold on to
electrons with enough strength to form much in the way of covalent bonds. For a covalent bond to form,
we need two atoms that both attract electrons with high electron affinity. Hence, the great majority of
covalent bonds will be formed between two non-metals. When both atoms in a bond are from the right
side of the periodic table, you can be sure that the bond is covalent. Here are some examples of molecules
with covalent bonds. You should check to see where the atoms involved in these molecules appear in the
periodic table. Covalent bonds form between atoms with relatively high electron affinity and they form
individual, separate molecules (Figure 11.6).

Phosphorus Trichloride

PCI,

:ci:P:ci:
":Cl: "

• •

:CI-P-Cl:
:CI:

H 2

)

«v<

H
:o:H

• •

:0-H

Carbon Dioxide

co„

:0:: C::o:

:o = C = o:

Figure 11.6: Various methods of showing a covalent bond.

The differences between ionic and covalent bonds are explained by the use of scientific models and examples
from nature. Video on Demand - The World of Chemistry - Chemical Bonds (http : //www . learner . org/
vod/vod_window . html?pid=800)

Multiple Bonds

So far, we have discussed covalent bonds in which one pair of electrons is shared. This type of bond is
called a single bond. Some atoms can share more than one pair of electrons. When atoms share two pairs
of electrons, it is called a double bond and when atoms share three pairs of electrons, it is called a triple
bond.

Double bonds are formed when atoms overlap two orbitals at the same time. An example of the formation

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336

of a double bond is in the bonding in the O2 molecule. Oxygen has six valence electrons and they are
distributed in the outermost energy level as 2s 2 2p 2 x 2p\2pl. Two of the p orbitals in the bonding shell of
oxygen are only half-filled and therefore are available for overlap. One of these half-filled p orbitals from
each oxygen atom will overlap end-to-end as shown in Figure 11.7.

Figure 11.7: The end-to-end overlap of orbitals forms a sigma bond.

Bonds formed by end-to-end overlap are called sigma (cr) bonds. The first bond formed between two atoms
is always a sigma bond. The second half-filled orbitals of the oxygen atoms will be oriented vertically and
cannot overlap end-to-end but they can overlap side-to-side as shown in the figure below.

The side-to-side overlap of orbitals forms a pi bond. (Source: CK-12 Foundation. CC-BY-SA)

The side-to-side overlap of orbitals forms a bond called a pi (n) bond. All the bonds formed between two
atoms after the first bond are pi bonds. The electron dot formula for a double bond would show two pairs
of electrons shared between atoms (see below figure) .

Oxygen is double bonded. (Source: Richard Parsons. CC-BY-SA)

Double bonds are stronger than single bonds. They are not exactly twice as strong; sometimes they are
more than twice as strong and sometimes they are less than twice as strong. Oxygen is a reactive element
and it is surprising that there is so much elemental oxygen in our atmosphere. The explanation for the
existence of so much elemental oxygen is that the double bond is very strong and it takes a great deal of
energy to break the double bond in oxygen so that the oxygen atoms could react with something else.

The nitrogen molecule, N2, is triple bonded (Figure 11.8). That means that the two nitrogen atoms share
three pairs of electrons. The first bond will be an end-to-end overlap (cr bond) and the other two bonds
will be side-to-side overlaps (n bonds). If the end-to-end overlap are the p x orbitals, then the side-to-side
overlaps will be the p y and p z orbitals.

Atomic nitrogen is quite reactive but molecular nitrogen is un-reactive. The reason nitrogen molecules not
do react readily is that the molecule is held together by an unusually strong bond - namely a triple bond.
Approximately 78% of our atmosphere is made up of nitrogen molecules.

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£3

:N::'N:

Figure 11.8: Nitrogen is triple bonded.

Lewis Formulas

What was called "electron dot formulas" when drawing them for individual atoms become "Lewis dot for-
mulas" or "Lewis structures" or "Lewis formulas" when drawing them for molecules. The Lewis structures
of a molecule show how the valence electrons are arranged among the atoms of the molecule. These repre-
sentations are named after G. N. Lewis. The rules for writing Lewis structures are based on observations
of thousands of molecules. From experiment, chemists have learned that when a stable compound forms,
the atoms usually have a noble gas electron configuration. Hydrogen forms stable molecules when it shares
two electrons (sometimes called the duet rule). Other atoms involved in covalent bonding typically obey
the octet rule. (Note: Of course, there will be exceptions.)

Rules for Writing Lewis Structures

Decide which atoms are bonded.

Count all the valence electrons of all the atoms.

Place two electrons between each pair of bonded atoms.

Complete all the octets (or duets) of the atoms attached to the central atom.

Place any remaining electrons on the central atom.

If the central atom does not have an octet, look for places to form double or triple bonds.

Example 1:

Write the Lewis structure for water, H2O.

Step 1: Decide which atoms are bonded.

Begin by assuming the hydrogen atoms are bonded to the oxygen atom. i.e. Assume the oxygen atom is
the central atom. H — O — H.

Step 2: Count all the valence electrons of all the atoms.

The oxygen atom has 6 valence electrons and each hydrogen has 1. The total is 8.

Step 3: Place two electrons between each pair of bonded atoms.

H : O : H

Step 4: Complete all the octets or duets of the atoms attached to the central atom.

The hydrogen atoms are attached to the central atom and hydrogen atoms require a duet of electrons and
those duets are already present.

Step 5: Place any remaining electrons on the central atom.

The total number of valence electrons is 8 and we have already used 4 of them. The other 4 will fit around
the central oxygen atom.

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h: o :h

Is this structure correct?

Are the total number of valence electrons correct? Yes

Does each atom have the appropriate duet or octet of electrons? Yes

Example 2:

Write the Lewis structure for carbon dioxide, C02-

Step 1: Decide which atoms are bonded.

Begin by assuming the carbon is the central atom and that both oxygen atoms are attached to the carbon.

Step 2: Count all the valence electrons of all the atoms.

The oxygen atoms each have 6 valence electrons and the carbon atom has 4. The total is 16.

Step 3: Place two electrons between each pair of bonded atoms.

O : C : O

Step 4: Complete all the octets or duets of the atoms attached to the central atom.

: o : c : o:
• #

Step 5: Place any remaining electrons on the central atom.

We have used all 16 of the valence electrons so there are no more to place around the central carbon atom.

Is this structure correct?

Is the total number of valence electrons correct? Yes

Does each atom have the appropriate duet or octet of electrons?

NO - each oxygen has the proper octet of electrons but the carbon atom only has 4 electrons. Therefore,
this is NOT correct.

Step 6: If the central atom does not have an octet, look for places to form double or triple bonds.

Double bonds can be formed between carbon and each oxygen atom.

Notice this time, each atom is surrounded by 8 electrons.
Example 3:

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Write the Lewis structure for ammonia, NH3.

Solution

The most likely bonding for this molecule is nitrogen as the central atom and each hydrogen bonded to
the nitrogen. Therefore, we can start by putting nitrogen in the center and put the three hydrogen atoms
around it.

The nitrogen atom has five valence electrons and each hydrogen atom has one so the total number of
valence electrons is 8.

The process for writing Lewis structures.

h n h h:n:h h:n:h

h ii ** ii

The next step is to put a pair of electrons between every bonded pair of atoms so we put a pair of electrons
between each of the hydrogen and nitrogen. The next step is to complete the octet or duet of each of
the non-central atoms. In this case, all the non-central atoms are hydrogen and they already have a duet
of electrons. The next step is to put all the left over electrons around the central atom. We have two
electrons left over so they would complete the octet for nitrogen. If the central atom, at this point, does
not have an octet of electrons, we would look for places to create a double or triple bond but in this case,
the central atom does have an octet of electrons. The final drawing on the right is the Lewis structure for
ammonia.

Example 4:

Given the skeleton structure for nitric acid, HNO3, place the electrons into a proper Lewis structure.

Solution

The skeleton for nitric acid has the three oxygen atoms bonded to the nitrogen and the hydrogen bonded
to one of the oxygen atoms. The total number of valence electrons is 5 + 6 + 6 + 6+1 = 24.

The process of writing the Lewis structure for nitric acid.

N O H 0:N:0:H

1 O 2 6

:6:N:6:H : 0::N:6'.H

3 O: 4 :0:

•• ••

The skeleton structure is given in "1" in the figure above. The next step is to put in a pair of electrons
between each bonded pair - this is done in "2." So far, we have accounted for 8 of the 24 valence electrons.
The next step is to complete the octet or duet for each of the non-central atoms. This is completed in "3."
At that point, we have used all of the valence electrons and the central atom does not have an octet of
electrons. The rules tell us to find a place to put a double or triple bond. For the amount of knowledge we
have at this point, any of the three oxygen atoms is just as good as the others for a double bond, so we
move two of the electrons around the far left oxygen atom and make a double bond between that oxygen
and the nitrogen. Now every atom in the molecule has its appropriate octet or duet of electrons. We have
a satisfactory Lewis structure for the nitric acid molecule.

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Lesson Summary

• Covalent bonds are formed between atoms with relatively high electron affinity.

• Some atoms are capable of forming double or triple bonds.

• Multiple bonds between atoms require multiple half-filled orbitals.

• End-to-end orbital overlaps are called sigma bonds.

• Side-to-side orbital overlaps are called pi bonds.

• Lewis structures are commonly used to show the valence electron arrangement in covalently bonded
molecules.

Review Questions

1. Which of the following compounds would you expect to be ionically bonded and which covalently
bonded?

Table 11.1:

Compound Ionic or Covalent

CS 2

K 2 S

FeF%

PF 3

BF 3

AIF 3

BaS

2. How many sigma bonds and how many pi bonds are present in a triple bond?

3. Draw the Lewis structure for CC/4.

4. Draw the Lewis structure for S02-

Further Reading / Supplemental Links

• http : //learner . org/resources/series61 . html

Vocabulary

covalent bond A type of bond in which electrons are shared by atoms.

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diatomic molecule A molecule containing exactly two atoms.

double bond A bond in which two pairs of electrons are shared.

triple bond A bond in which three pairs of electrons are shared.

sigma bond A covalent bond in which the electron pair is shared in an area centered on a line running
between the atoms.

pi bond A covalent bond in which p orbitals share an electron pair occupying the space above and below
the line joining the atoms.

11.3 Naming Covalent Compounds

Lesson Objectives

• The student name covalent compounds using the IUPAC nomenclature system.

• The student will provide formulas for covalent compounds given the IUPAC name.

Introduction

The systematic procedure for naming chemical compounds uses a different approach for different types of
compounds. In a previous chapter, we have discussed the procedures for naming binary ionic compounds,
ionic compounds involving polyatomic ions, and ionic compounds involving metals with variable oxidation
states. In this section, we will describe the system used for covalently bonded compounds. Because of the
large numbers of covalent compounds that may form between the same two elements, the nomenclature
system for covalent compounds is quite different.

The Number of Atoms in the Formulas Must be Indicated

In naming ionic compounds, there is no need to indicate the number of atoms of each element in a
formula because, for most cases, there is only one possible compound that can form from the ions present.
When aluminum combined with sulfur, the only possible compound is aluminum sulfide, AI2S3. The only
exception to this is a few variable oxidation number metals and those are handled with Roman numerals
for the oxidation number of the metal, as in iron (II) chloride, FeCli-

With covalent compounds, however, we have a very different situation. There are six different covalent
compounds that can form between nitrogen and oxygen and in two of them, nitrogen has the same oxidation
number. Therefore, the Roman numeral system will not work. Chemists devised a nomenclature system
for covalent compounds that indicate how many atoms of each element is present in a molecule of the
compound.

Greek Prefixes

In naming binary covalent compounds, four rules apply:

1. The first element in the formula is named first using the normal name of the element.

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2. The second element is named as if it were an anion. Note: There are no ions in these compounds but
we use the "-ide" ending on the second element as if it were an anion.

3. Greek prefixes are used for each element to indicate the number of atoms of that element present in the
compound.

Table 11.2: Greek Prefixes

Prefix

Number Indicated

Mono-

Di-

Tri-

Tetra-

Penta-

Hexa-

Hepta-

Octa-

Nona-

Deca-

1
2
3
4
5
6
7
8
9
10

4. The prefix "mono-" is never used for naming the first element. For example, CO is called carbon
monoxide, not monocarbon monoxide.

Examples

N 2
NO

N0 2

N 2 3
N 2 A
N 2 5

co 2

PaOiq

P2S 5

dinitrogen monoxide
nitrogen monoxide
nitrogen dioxide
dinitrogen trioxide
dinitrogen tetraoxide
dinitrogen pentaoxide
sulfur hexafluoride
carbon dioxide
tetraphosphorus decaoxide
diphosphorus pentasulfide

Lesson Summary

• Covalently bonded molecules use Greek prefixes in their nomenclature.

Review Questions

1. Name the compound CO.

2. Name the compound PCI3.

3. Name the compound PC/5.

4. Name the compound N 2 0-$.

5. Name the compound BCI3.

6. Name the compound SF4.

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7. Name the compound ChO.

8. Write the formula for the compound sulfur trioxide.

9. Write the formula for the compound dinitrogen tetrafluoride.

10. Write the formula for the compound oxygen difluoride.

11. Write the formula for the compound dinitrogen pentoxide.

12. Write the formula for the compound sulfur hexafluoride.

13. Write the formula for the compound tetraphosphorus decaoxide.

Further Reading / Supplemental Links

• http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp/bomls2_ll .swf

• http : //www . visionlearning . com/library /module_viewer . php?mid=55

Image Sources

(1) CK-12 Foundation. . CC-BY-SA.

(2) Richard Parsons. . CC-BY-SA.

(3) Richard Parsons. . CC-BY-SA.

(4) Richard Parsons. . CC-BY-SA.

(5) Richard Parsons. . CC-BY-SA.

(6) Richard Parsons. . CC-BY-SA.

(7) CK-12 Foundation. . CC-BY-SA.

(8) Richard Parsons. Nitrogen is triple bonded.. CC-BY-SA.

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Chapter 12

Reactions

12.1 Chemical Equations

Lesson Objectives

The student will read chemical equations and provide requested information contained in the equation

including information about substances, reactants, products, and physical states.

The student will convert symbolic equations into word equations and vice versa.

The student will use the common symbols, — >,+, (s)(l)(g)(aq) appropriately.

The student will describe the roles of subscripts and coefficients in chemical equations.

The student will balance chemical equations with the simplest whole number coefficients.

Introduction

A chemical change occurs when some substances (it could even be the same substance) come into contact,
the chemical bonds of the substances break, and the atoms that compose the compounds separate and
re-arrange themselves into new compounds with new chemical bonds. When this process occurs, we call it
a chemical reaction. In order to describe a chemical reaction, we need to indicate what substances were
present at the beginning and what substances were present at the end. The substances that were present
at the beginning are called reactants and the substances present at the end are called products. In many
chemical reactions, it is necessary not only to name the reactants and products but also to indicate the
phase of each substance.

Molecules not only have a particular group of atoms arranged in an exact way, they also have a specific
amount of potential energy associated with their chemical bonds. When reactants re-arrange the orga-
nization of their atoms and bonds, the amount of potential energy associated with the reactant bonds is
almost never the same as the amount of energy associated with the product bonds. Therefore, energy must
be absorbed or given off by the reaction. For some reactions, the potential energy difference between the
reactant bonds and product bonds must be represented in the equation.

All of this information about a chemical reaction, reactants, products, phases, and energy changes can be
described in words or it can be presented by a special shorthand notation devised precisely for communi-
cating this information.

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Some Interactions Represent Chemical Reactions

We know already that chemistry is the study of matter and that the Law of Conservation of Matter states
that matter is conserved in a chemical reactions. But what does this really mean? When you place an ice
cube on the counter and it melts, do you think the mass has changed? Of course you don't, but why not?
The reason is because the mass before and after a (non-nuclear) change has to remain the same. Energy
works the same way. The Law of Conservation of Energy states that total energy involved in a chemical
reaction is conserved from reactants to products.

Energy cannot be created nor destroyed in a closed system. Think of it as balancing on a seesaw (Figure
??). If energy is absorbed by the reaction, it has to have come from somewhere in the closed system.

Let's look at an example:

H 2 {s) + 3312 J -» H 2 (L)

Thus it takes 3312 Joules of energy to melt one ice cube of approximately 0.01 kg. The 3312 Joules is
an energy unit. Where did the energy come from? The energy came from the kinetic energy in the room
where the ice cube was melting. So the energy was absorbed by the ice cube and released by the room.
Therefore, the total amount of energy remained constant in the room. The mass of the ice cube is the
same as the mass of the liquid puddle. Therefore mass was conserved. See how it works?

We will go into further depth with this concept as we explore chemical reactions in detail through the
remainder of this unit.

Reactants Change into Products

Sometimes when reactants are put into a reaction vessel, a reaction will take place to produce products.
Reactants are the starting materials, that is, whatever we have as our initial ingredients. The products are
just that, what is produced or the result of what happens to the reactants when we put them together in
the reaction vessel. If we think about baking chocolate chip cookies, our reactants would be flour, butter,
sugar, vanilla, some baking soda, salt, egg, and chocolate chips. What would be the products? Cookies!
The reaction vessel would be our mixing bowl.

Flour + Butter + Sugar + Vanilla + Baking Soda + Eggs + Chocolate Chips — » Cookies

Chemical reactions are similar. If sulfur dioxide is added to oxygen, sulfur trioxide is produced. Sulfur
dioxide and oxygen (SO2 + O2) are reactants and sulfur trioxide (SO3) is the product.

2 S0 2 ( g ) +■ 2(g ) ► 2 S0 3 |g)

Reactants Products

In chemical reactions, the reactants are found before the symbol "— >" and the products and found after
the symbol "— »". The general equation for a reaction is:

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Reactants — » Products
We will explore more about this in the next section.

Symbol Equations are Shorthand for Word Equations

There are a few special symbols that we need to know in order to "talk" in chemical shorthand. Remember
when you started learning pre-algebra and you learned the order of operations. We may have all learned
the acronym BEDMAS for brackets, exponents, division, multiplication, addition, and subtraction. The
acronym was a shorthand method or symbolic notation for the order of operations when we want to "talk"
in mathematical equations. We have the same thing for chemistry when we want to "talk" in chemical
equations. In the table below is the summary of the major symbols used in chemical equations. You will
find there are others but the main ones that we need to memorize are listed in Table 12.1.

Table 12.1: Common Symbols

Symbol Meaning

— » Symbol used to separate reactants and products;

means "to produce", can be read as "yields"
Ex:2 H 2 + 2 -» 2 H 2

+ Symbol used to separate reactants and/or products

in a chemical reaction; means "is added to".
2H 2 + 2 -» 2 H 2
AgN0 3 + NaCl -> AgCl + NaN0 3

(s) In the solid state Sodium in the solid state: Na(s)

Gold in the solid state: Au(s)

(I) In the liquid state Water in the liquid state: H20{1)

Mercury in the liquid state: Hg(l)

(g) In the gaseous state Helium in the gaseous state:

He{g)
Carbon dioxide in the gaseous state: C 02(g)

(aq) In the aqueous state, dissolved in water to make a

solution Sodium chloride solution: NaCl(aq)
Hydrochloric acid solution: HCl(aq)

Chemists have a choice of methods for describing a chemical reaction. They could draw a picture of the
chemical reaction.

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■ ^

Or, they could write a word equation for the chemical reaction.

Two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of water
vapor.

Or, they could write the equation in chemical shorthand.

2 H 2 (g) + 2(g) -» 2 H 2 {g)

In the symbolic equation, chemical formulas are used instead of chemical names for reactants and products
and symbols are used to indicate the phase of each substance. It should be apparent that the chemical
shorthand method is the quickest and clearest method for writing chemical equations.

I could write that an aqueous solution of calcium nitrate is added to an aqueous solution of sodium
hydroxide to produce solid calcium hydroxide and an aqueous solution of sodium nitrate. Or in shorthand
I could write:

Ca(N0 3 ) 2{aq) + 2 NaOH {aq) -» Ca{OH) 2{s) + 2 NaN0 3(aq)

How much easier is that to read? Let's try it in reverse? Look at the following reaction in shorthand
notation and write the word equation for the reaction.

Cu (s) + AgN0 3iaci} -> Cu(N0 3 ) 2 (aq) + Ag( s )

The word equation for this reaction might read something like "solid copper reacts with an aqueous solution
of silver nitrate to produce a solution of copper (II) nitrate with solid silver".

Sample question: Transfer the following symbolic equations into word equations or word equations into
symbolic equations.

(a)HCl {aq) +NaOH {aq) -> NaCl (aq) + H 2 {L)

(b) Gaseous propane (C 3 Hg) burns in oxygen gas to produce gaseous carbon dioxide and liquid water.

(c) Hydrogen fluoride gas reacts with an aqueous solution of potassium carbonate to produce an aqueous
solution of potassium fluoride, liquid water, and gaseous carbon dioxide.

Solution:

(a) An aqueous solution of hydrochloric acid reacts with an aqueous solution of sodium hydroxide to
produce an aqueous solution of sodium chloride and liquid water.

(b) C 3 // 8(g) + 2 (g) -> c °2( g ) + H 2 0( L )

Lesson Summary

• Chemical reactions can be represented by word equations or by formula equations.

• Formula equations have reactants on the left, an arrow that is read as "yields," and the products on
the right.

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Review Questions

1. Mothballs are commonly used to preserve clothing in "off-season." We recognize mothballs due to
its smell because of a chemical compound known as Naphthalene, CiqHs- What are the different
elements found in naphthalene and how many atoms of each are found in the formula?

2. Do you think a chemical reaction occurs every time two substances are placed together in a reaction
vessel?

3. Transfer the following symbolic equations into word equations.

(a) H 2 S0 4{aq) +NaCN {aq) -> HCN {aq) + Na 2 S0 4{aq)

(b) Cu(s) + AgN0 3{aq) -» Ag {s) +Cu(N0 3 ) 2{aq]

4. Transfer the following equations from word equations into symbolic equations.

(a) Solid calcium metal is placed in liquid water to produce aqueous calcium hydroxide and hydrogen
gas.

(b) Gaseous sodium hydroxide is mixed with gaseous chlorine to produce aqueous solutions of sodium
chloride and sodium hypochlorite plus liquid water.

5. Did you know that you can simulate a volcanic eruption in a lab that looks like the real thing? A
source of heat is gently placed it into a mound of ammonium dichromate. The ammonium dichromate
decomposes to solid chromium (III) oxide, nitrogen monoxide gas, and water vapor. Write the
symbolic reaction for the "volcanic eruption".

Vocabulary

reactants The starting materials in the reaction (i.e. S0 3 + H 2 — » H 2 SO^;S0 3 + H 2 = reactants).

products The materials present at the end of a reaction (i.e. S 3 + H 2 — > H 2 SO^H 2 SO^= product).

— » Symbol used to separate reactants and products; means to produce. Ex: 2 H 2 + 2 — » 2 H 2 0.

+ Symbol used to separate reactants and/or products in a chemical reaction; means is added to. Examples:
2// 2 + 2 -» 2 H 2 0;AgN0 3 + NaCl -» AgCl + NaN0 3 .

(s) In the solid state. Examples: Sodium: Na(s); Gold: Au(s).

(I) In the liquid state. Examples: Water: H 2 0(l); Mercury: Hg(l).

(g) In the gaseous state. Examples: Helium: He(g); Carbon dioxide: C 02(g)-

(aq) In the aqueous state. Examples: Sodium chloride solution: NaCl(aq).

12.2 Balancing Equations

Lesson Objectives

Demonstrate the Law of Conservation of Matter in a chemical reaction.
Explain the roles of coefficients and subscripts in a chemical reaction.
Balance equations using the simplest whole number coefficients.

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Introduction

Even though chemical compounds are broken up and new compounds are formed during a chemical reaction,
atoms in the reactants do not disappear nor do new atoms appear to form the products. In chemical
reactions, atoms are never created or destroyed. The same atoms that were present in the reactants are
present in the products - they are merely re-organized into different arrangements. In a complete chemical
equation, the two sides of the equation must be balanced. That is, in a complete chemical equation, the
same number of each atom must be present on the reactants and the products sides of the
equation.

Subscripts and Coefficients

There are two types of numbers that appear in chemical equations. There are subscripts which are part
of the chemical formulas of the reactants and products and there are coefficients that are placed in front
of the formulas to indicate how many molecules of that substance is used or produced.

Coefficients
Cu (s) + 2 AgN0 3 (a q ) — >Cu(N0 3 ) 2(aq) + 2 Ag (s)

Subscripts

The subscripts are part of the formulas and once the formulas for the reactants and products are determined,
the subscripts may not be changed. The coefficients indicate how many molecules of each substance is
involved in the reaction and may be changed in order to balance the equation. The equation above indicates
that one atom of solid copper is reacting with two molecules of aqueous silver nitrate to produce one
molecule of aqueous copper (II) nitrate and two atoms of solid silver. When you learned how to write
formulas, it was made clear that when only one atom of an element is present, the subscript of "1" is not
written - so that when no subscript appears for an atom in a formula, you read that as one atom. The
same is true in writing balanced chemical equations. If only one atom or molecule is present, the coefficient
of "1" is omitted.

Coefficients are inserted into the chemical equation in order to balance it; that is, to make equal the total
number of each atom on the two sides of the equation. Consider the equation representing the reaction
that occurs when gaseous methane, C//4, is burned in air (reacted with oxygen gas), and produces gaseous
carbon dioxide and liquid water.

The unbalanced symbolic equation for this reaction is given below.

Ci?4(g) + 2 (g) -* C0 2 { g ) + H 2 0( L ) Equation 1

It is quickly apparent that equation 1 is not balanced. There are 4 hydrogen atoms in the reactants and
only 2 hydrogen atoms in the products. The oxygen atoms in the equation are also not balanced. In order
to balance this equation, it is necessary to insert coefficients in front of some of the substances to make the

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numbers of atoms of each element the same on the two sides of the equation. We can begin this process
by placing a coefficient of 2 in front of the water molecule.

11 A

C// 4 ( g ) + 2 (g) -> C0 2 ( g ) + 2 H 2 0( L ) Equation 2

The insertion of this coefficient balanced the hydrogen atoms (there are now 4 on each side) but the
equation is still not completely balanced. The oxygen atoms are not balanced. There are 4 oxygen atoms
in the products but only two in the reactants. We can now insert a coefficient of 2 in front of the oxygen
molecule in the reactants.

•

C// 4 ( g ) + 2 2 ( g ) -» C0 2 ( g ) + 2 H 2 0( L ) Equation 3

When we count up the number of each type of atom on the two sides of the equation now, we see that the
equation is properly balanced. There is one carbon atom on each side, four hydrogen atoms on each side,
and four oxygen atoms on each side. Equation balancing is accomplished by changing coefficients, never
by changing subscripts.

The Process of Balancing an Equation

The process of writing a balanced chemical equation involves three steps. As a beginning chemistry student,
you will not know whether or not two given reactants will react or not and even if you saw them react,
you would not be sure what the products are without running tests to identify them. Therefore, for the
time being, you will be told both the reactants and products in any equation you are asked to balance.

Step 1
Step 2
Step 3

Know what the reactants and products are, and write a word equation for the reaction.
Write the formulas for all the reactants and products.
Adjust the coefficients to balance the equation.

You must keep in mind that there are a number elements commonly appearing in equations that, under
normal conditions, exist as diatomic molecules (Table 12.2).

Table 12.2: Elements that exist as diatomic molecules under normal conditions

Element Formula for Diatomic Molecule Phase Under Normal Conditions

Hydrogen H 2 Gaseous

Oxygen 2 Gaseous

Nitrogen N 2 Gaseous

Chlorine C/2 Gaseous

Fluorine F 2 Gaseous

Bromine Br 2 Liquid

Iodine I 2 Solid

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When the names of any of these elements appear in word equations, you must remember that the name
refers to the diatomic molecule and insert the diatomic formula into the symbolic equation. If, under
unusual circumstances, it was desired to refer to the individual atoms of these elements, the text would
refer specifically to atomic hydrogen or atomic oxygen and so on.

Sample Problem 1

Write a balanced equation for the reaction that occurs between chlorine gas and aqueous sodium bromide
to produce liquid bromine and aqueous sodium chloride.

Step 1: Write the word equation (keeping in mind that chlorine and bromine refer to the diatomic
molecules).

Chlorine + sodium bromide — » bromine + sodium chloride

Step 2: Substitute the correct formulas into the equation.

Cl 2 + NaBr -> Br 2 + NaCl

Step 3: Insert coefficients where necessary to balance the equation.

By placing a coefficient of 2 in front of the NaBr, we can balance the bromine atoms and by placing a
coefficient of 2 in front of the NaCl, we can balance the chloride atoms.

Cl 2 + 2 NaBr -> Br 2 + 2 NaCl

A final check (always do this) shows that we have the same number of each atom on the two sides of the
equation and we do not have a multiple set of coefficients so this equation is properly balanced.

Sample Problem 2

Write a balanced equation for the reaction between aluminum sulfate and calcium bromide to produce
aluminum bromide and calcium sulfate. It might be worthwhile to note here that the reason you memorized
the names and formulas for the polyatomic ions is that these ions usually remain as a unit throughout
many chemical reactions.

Step 1: Write the word equation.

Aluminum sulfate + calcium bromide — » aluminum bromide + calcium sulfate

Step 2: Replace the names of the substances in the word equation with formulas.

Al 2 (S A ) 3 + CaBr 2 -» AlBr 3 + CaS A Equation 1

Step 3: Insert coefficients to balance the equation.

In order to balance the aluminum atoms, we must insert a coefficient of 2 in front of the aluminum
compound in the products.

Al 2 (S 04)3 + CaBr 2 -> 2 AlBr 3 + CaS 4 Equation 2

In order to balance the sulfate ions, we must insert a coefficient of 3 in front of the CaSO^ in the products.
Al 2 (S0 4 ) 3 + CaBr 2 -» 2 AlBr 3 + 3 CaSO± Equation 3

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In order to balance the bromine atoms, we must insert a coefficient of 3 in front of the CaBr2 in the
reactants.

Ah{S0 4 )3 + 3 CaBr 2 -> 2 AlBr 3 + 3 CaS0 4 Equation 4

The insertion of the 3 in front of the CaBr2 in the reactants also balances the calcium atoms in the CaS O4
in the products. A final check shows 2 aluminum atoms on each side, 3 sulfur atoms on each side, 12 oxygen
atoms on each side, 3 calcium atoms on each side, and 6 bromine atoms on each side. This equation is
balanced.

Sample Problem 3

Balance the following skeletal equation. (The term "skeletal equation" refers to an equation that has the
correct formulas but has not yet had the proper coefficients added.)

Fe(N0 3 ) 3 + NaOH -> Fe(OH) 3 + NaN0 3 (skeletal equation)

We can balance the hydroxide ion by inserting a coefficient of 3 in front of the NaOH on the reactant side.
Fe{N0 3 ) 3 + 3 NaOH -> Fe(OH) 3 + NaN0 3 (intermediate equation)

Then we can balance the nitrate ions by inserting a coefficient of 3 in front of the sodium nitrate on the
product side.

Counting the number of each type of atom on the two sides of the equation will now show that this equation
is balanced.

Simplest Whole Number Coefficients

Chemical equations should be balanced with the simplest whole number coefficients that balance the
equation. Here is the properly balanced equation from the previous section.

AZ 2 (S 6*4)3 + 3 CaBr 2 -» 2 AlBr 3 + 3 CaSOt Equation 4

You should note that the equation in the previous section would have the same number of atoms of each
type on each side of the equation with the following set of coefficients.

2 Al 2 {S0 4 ) 3 + 6 CaBr 2 -» 4 AlBr 3 + 6 CaS0 4 Equation 5

You should count the number of each type of atom on each side of the equation to confirm that this
equation is "balanced". While this set of coefficients do "balance" the equation, they are not the lowest
set of coefficients possible that balance the equation. We could divide each of the coefficients in this
equation by 2 and get a another set of coefficients that are whole numbers and also balance the equation.
Since it is required that an equation be balanced with the lowest whole number coefficients, equation 5 is
NOT properly balanced. The properly balanced equation for this reaction is equation 4. When you have
finished balancing an equation, you should not only check to make sure it is balanced, you should also
check to make sure that it is balanced with the simplest set of whole number coefficients possible.

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Let's consider another equation for balancing; the combustion of octane, C 8 His-

C 8 H 18 + 2 ^> C0 2 + H 2 Equation 1

To begin balancing this equation, we note that there are 8 carbon atoms on the reactant side and only 1
carbon on the product side. We can balance the carbon atoms by inserting a coefficient of 8 in front of the
carbon dioxide on the product side.

C 8 H i8 + <9 2 -> 8 C0 2 + H 2 Equation 2

Next, we note that there are 18 hydrogen atoms in the reactants and only 2 hydrogen atoms in the products.
We can balance the hydrogen atoms by inserting a coefficient of 9 in front of the water molecule on the
product side.

C 8 H 18 + 2 -> 8 C0 2 + 9 H 2 Equation 3

Now when we count up the number of oxygen atoms in the products, we find there are 25 oxygen atoms.
What coefficient can you insert in front of the 2 molecule in the reactants to have 25 oxygen atoms? Since
the oxygen molecule has a subscript of 2, the only coefficient for this molecule that will produce 25 oxygen
atoms is a coefficient of y. If we insert y as the coefficient for the oxygen molecule, we get a balanced
equation but we do not get an equation that is balanced with the simplest whole number coefficients.

25
C 8 H 18 + — 2 -> 8 C0 2 + 9 H 2 Equation 4

In order to have whole number coefficients for this equation, we must go through the equation and multiply
all the coefficients by 2. That produces the coefficients shown in equation 5.

2 C 8 H 18 + 25 2 -» 16 C0 2 + 18 H 2 Equation 5

Now the equation is balanced with the simplest whole number coefficients and is a properly balanced
equation.

Sample Problem

Given the following skeletal (un-balanced) equations, balance them.

(a) CaC0 3 ^ — > CaOi s \ + C0 2 r g \

(b) H 2 SO i{aq) +Al{OH) 3(aq) ^Al 2 {SO A ) 3{aq) +H 2 {L)

(d) C 2 // 6(g ) + 2 (g) -» C0 2 ( S ) + H 2 0( L )
Solutions

(a) CaC0 3 ( s ) — » CaOi s \ + C0 2 t g \ (sometimes the equation balances with all coefficients of 1)

(b) 3 H 2 SO A(aq) + 2 Al{OH) 3{aq) -> Al 2 (S0 4 ) 3{aq) + 6 H 2 {L)

(d) 2 C 2 H 6{g) + 7 2{g) -» 4 C0 2(g) + 6 H 2 {L)

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The Conservation of Mass in Chemical Reactions

Both the numbers of each type of atom and the mass are conserved during chemical reactions. An exam-
ination of a properly balanced equation will demonstrate that mass is conserved. Consider the following
reaction.

Fe(N0 3 ) 3 + 3 NaOH -> Fe{OH) 3 + 3 NaN0 3

You should demonstrate that this equation is balanced by counting the number of each type of atom on
each side of the equation.

We can also demonstrate that mass is conserved in this reaction by determining the total mass on the two
sides of the equation.

Fe{N0 3 ) 3 + 3 NaOH -^ Fe(OH) 3 + 3 NaN0 3

Reactant Side Mass

1 molecule of Fe{N0 3 ) 3 x molecular weight = (1) (241.9 daltons) = 241.9 daltons

3 molecules of NaOH x molecular weight = (3) (40.0 daltons) = 120. daltons

Total mass for reactants = 241.9 daltons + 120. daltons = 361.9 daltons

Product Side Mass

1 molecule of Fe{OH) 3 x molecular weight = (1)( 106.9 daltons) = 106.9 daltons

3 molecules of NaN0 3 x molecular weight = (3) (85.0 daltons) = 255 daltons

Total mass for products = 106.9 daltons + 255 daltons = 361.9 daltons.

As you can see, both number of atom types and mass are conserved during chemical reactions. A group of
20 objects stacked in different ways will still have the same total mass no matter how you stack them.

Lesson Summary

• To be useful, chemical equations must always be balanced.

• Balanced chemical equations must have the same number and type of each atom on both sides of the
equation.

• The coefficients in a balanced equation must be the simplest whole number ratio.

• Mass is always conserved in chemical reactions.

Review Questions

1. Explain in your own words why it is essential that subscripts remain constant but coefficients can
change.

2. Which set of coefficients will properly balanceLthe following^ equation?

(a) 1,1,1,1

(b) 1,3,2,2

(d) 2,7,4,6

3. When properly balanced, what is the, sum oXall thqcoefficients^in the following chemical equation?

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(a) 4

(b) 7

(d) None of the above

4. When the following equation is balanced, what is the coefficient found in front of the 2 ?

Pi + 2 + H 2 -» H 3 POi

(a) 1

(b) 3

(d) 7

5. Balance the following equations.

(a) XeF 6(s) + H 2 {L) -» XeO s(s) + HF (g)

(b) C« w + AgN0 3(aq) -> Ag {s) + CM(W0 3 )2( a? )

(d) A7(0#) 3 + Mg 3 ( J P0 4 )2 -» A/P0 4 + Mg{OH) 2

Further Reading / Supplemental Links

Website with lessons, worksheets, and quizzes on various high school chemistry topics.

• Lesson 8-1 is on Balancing Equations.

• http : //www . f ordhamprep . org/gcurran/sho/sho/lessons/lesson81 . htm

Vocabulary

law of conservation of matter Matter is neither created nor destroyed in chemical reactions.

skeletal equation a chemical equation before it has been balanced

balanced chemical equation a chemical equation in which the number of each type of atom is equal
on the two sides of the equation

12.3 Types of Reactions

Lesson Objectives

• Identify the types of reactions.

• Predict the products in different types of reactions.

• Distinguish between the different types of reactions.

• Write balanced chemical equations and identify the reaction type given only the reactants.

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Introduction

Chemical reactions are classified into groups to help us analyze them and also to help us predict what the
products of the reaction will be. The four major types of chemical reactions are synthesis, decomposition,
single replacement, and double displacement. Sometimes other names are used for these basic types of
reactions but the same four are always listed. There are also some sub-groups under these four but we will
concentrate on the basic four groups.

Synthesis Reactions

A synthesis reaction is one in which two or more reactants combine to make ONE type of product.

General equation: A + B — » AB

Synthesis reactions occur as a result of two or more simpler elements or molecules combining to form a more
complex molecule. Look at the example below. Here two elements (hydrogen and oxygen) are combining
to form one product (water).

Example: 2 H 2 (g) + 2 { g ) -> 2 H 2 0( L ) + energy

(notice that energy is a product and therefore this synthesis reaction is exothermic)

We can always identify a synthesis reaction because there is only one product or one substance present
after the arrow (on the right hand side) of the reaction.

We can write the synthesis reaction for sodium chloride just by knowing the elements that are present in
the product.

2 Na(s) + Cl 2 (g) -> 2 NaCl(s)

You will recognize synthesis reactions by the fact that there is only one product. You should be able to
write a formation synthesis reaction if you are given a product by picking out its elements and writing the
reaction (just like we did with NaCl). Also, if you are given elemental reactants and told that the reaction
is a synthesis reactant, you should be able to predict the products.

Sample Question:

(a) Write a synthesis reaction to produce silver bromide, AgBr.

(b) Predict the products for the following reaction: C0 2 ^ + H 2 Ot L \ — »

(a) 2 Ag (s) + Br 2{L) -» 2 AgBr (s)

(b) C0 2{g) + H 2 {L) -» H 2 C0 3H)

Decomposition Reactions

The best way to remember a decomposition reaction is that for all reactions of this type, there is only one
reactant. When ONE type of reactant breaks down to form two or more products, we have a decomposition
reaction.

General Equation: AB — > A + B

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Look at the example below for the decomposition of ammonium nitrate to dinitrogen oxide and water.

Example: energy + NH A NO^ s ) -» N 2 0( g ) + 2 H 2 0( g )

(Notice the energy is a reactant and therefore this decomposition reaction is endothermic)

Notice the one reactant jNH/jNO^a, is on the left of the arrow and there is more than one on the right side
of the arrow. This is the exact opposite of the synthesis reaction type.

When studying decomposition reactions, we can predict reactants in a similar manner as we did for synthesis
reactions. Look at the formula for magnesium nitride, Mg 3 N 2 . What elements do you see in this formula?
You see magnesium, Mg, and nitrogen, N 2 . Now we can write a decomposition reaction for magnesium
nitride.

Mg 3 N 2{s) -> 3 Mg {s) + N 2 ( g )

Notice there is only one reactant.
Sample Question:

(a) Write a decomposition reaction for aluminum oxide, Al 2 3 .

(b) Predict the products for the following reaction: Ag 2 S — >

(a) 2 Al 2 3 -»4 Al + 3 2

(b) Ag 2 S -^2Ag + S

Single Replacement (metal)

A third type of reaction is the single replacement reaction. In single replacement reactions one element
reacts with one compound to form products. The single element is said to replace an element in the
compound when products form, hence the name single replacement.

There are actually three different types of single replacement reactions; 1) the single element is a metal and
replaces the metal of the compound in the second reactant, 2) the single element is a metal and replaces
the hydrogen of the compound in the second reactant which is always an acid, and 3) the single element
is a non-metal and replaces the non-metal in the compound. The single replacement reaction in which
hydrogen is replaced in an acid will be covered in detail in a later section. In this section, let us focus
on single replacement reactions where the element reactant replaces the metal in the compound reactant.
Look at the general reaction below.

General equation: A + BC — » B + AC

(element replaces cation of compound)

Example: Zn {s) + Cu(N0 3 ) 2 ( aq ) -» Zn{N0 3 ) 2 ( aq) + Ci4( s )

Notice there is only one reactant that is an element and one reactant that is a compound.

When studying single replacement reactions, we can predict reactants in a similar manner as we did for
synthesis and decomposition reactions. Remember the oxidation numbers learned previously. This prior
knowledge will become especially important here when predicting products for the rest of this chapter.
Let's try an example where we will predict a series of products for an element reacting with a compound
where the element will replace the metal ion.

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Suppose that we know a single replacement reaction will occur between solid aluminum and solid iron (III)
oxide.

A/ w + Fe 2 Hs) ->

In order to predict the products we need to know that aluminum will replace iron and form aluminum
oxide. Aluminum has a charge of +3 (it is in group 3A) and oxygen has a charge of -2 (it is in group 6A).
The compound formed between aluminum and oxygen, therefore, will be Al-zO^uy Since iron is replaced
in the compound by aluminum, the iron will now be the single element in the products. The unbalanced
equation will be:

Al {s) + Fe 2 3{s) -> Al 2 3(s) + Fe {s)

and the balanced equation will be:

2 A/ w + Fe 2 3{s) -» Al 2 3{s) + 2 Fe (s)

Sample Question:

(a) Write a single replacement reaction for the reaction between zinc solid and lead (II) nitrate solution to
produce zinc nitrate solution and solid lead.

(b) Predict the products for the following reaction: Fe + CuS O4 — >
(b) Predict the products for the following reaction: AI + CuCl 2 — >
Solution:

(a) Zn {s) + Pb(N0 3 ) 2(aq) -> Pb (s) + Zn(N0 3 ) 2(aq)

(b) Fe {s) + CuSO i{aq) -> FeS0 4{aq) + Cu {s)

Single Replacement (many metals with acid)

These reactions are the same as those studied in the last section except the compound in the reactant
side of the equation is always an acid. Since you may not have studied acids yet, you should consider an
acid to be a compound in which hydrogen is combined with a anion. Therefore, in this section, the single
replacement reactions will be where the element reactant replaces the hydrogen in the compound reactant.
Look at the general reaction below.

General equation: A + 2 HC — » AC 2 + H 2 (element replaces hydrogen of compound)

Example: Zn {s) + 2 HBr {aq) -» ZnCl 2{aq) + # 2 (g)

When studying these single replacement reactions, we can predict reactants in a similar manner as we did
for the other types of single replacement reactions. Look at the reaction below. Since HCl is a compound
that has hydrogen combined with an anion, it is an acid.

Mg {s) + 2 HCl (aq) -»

In order to predict the products, we need to know that magnesium will replace hydrogen and form mag-
nesium chloride. Magnesium has a charge of +2 (it is in group 2) and chlorine has a charge of -1 (it
is in group 7A). Therefore, the compound formed will be MgCl 2 . When hydrogen is replaced from the

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compound, it will appear in the products as elemental hydrogen, H 2 . All we need to do after the products
are determined is to balance the equation.

Mg(s) + 2 HCl(aq) -> MgCl 2 (s) + H 2 (g)

Notice there is one reactant that is an element (Mg) and one reactant that is an acid compound. The Mg
has replaced the hydrogen in the HCl in the same manner as the Zn replaced the hydrogen in the HBr in
the example above.

Sample Question:

(a) Write a single replacement reaction for the reaction between iron solid and hydrochloric acid solution
to produce iron(II) chloride solution and hydrogen gas.

(b) Predict the products for the following reaction: Znt s \ + H 2 S 0^ aq ) — »

(a)Fe w + 2 HCl {aq) -> FeCl 2{aq) + H 2{g)
(b)Zn( s ) + H 2 S 0^ aq ) -» ZnS 0^ aq) + H 2 ( g )
(c)2A/ w + 6 HN0 3{aq) -» 2 A/(iV0 3 ) 3 W) + 3 H 2{g)

Single Replacement (non-metal)

In this section, we will focus on the final type of single replacement reactions where the element reactant
replaces the non-metal (or the anion) in the compound reactant. Look at the general reaction below.

General equation: A + BC — » C + BA (element replaces anion of compound)

Example: Cl 2 ( g ) + 2 Kl^ aq) -> 2 KCl {aq) + l 2 ( s )

Notice, in the example, the chlorine replaced the iodine to produce solid iodine as a product. We can predict
the products for these single replacement reactions in a similar manner as for all other single replacement
reactions. The only difference here is that we have to remember that we are replacing the anion of the
compound rather than the cation. Look at the reaction below between chlorine gas and sodium bromide.
This is an actual extraction method for extracting bromine from the ocean water found to contain sodium
bromide. Can you complete the reaction?

Cl 2{g) + NaBr {aq) ->

In order to predict the products of this reaction we need to know that chlorine will replace bromine and
form sodium chloride. Sodium has a charge of +1 (it is in group 1) and chlorine has a charge of -1 (it is
in group 17). The compound formed will be NaCl.

Cl 2 ( g ) + 2 NaBr {aq) -> 2 NaCl {aq) + Br 2{L)

Notice as with all of the other single replacement reactions, the reactants include one element and one
compound and the products contain one element and one compound. This is the determining factor for
identifying whether you have a single replacement reaction.

Sample Question:

(a) Write a single replacement reaction for the reaction between sodium iodide solution and bromine
solution to produce sodium bromide solution and solid iodide.

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(b) Predict the products for the following reaction: Br 2 ( aq ) + Kh aq \ — >
(b) Predict the products for the following reaction: Mgl 2 ^ + Cl^ig) — »
Solution:

(a) 2 Nal {aq) + Br 2{L) -> 2 NaBr (aq) + / 2(s)

(b) 5r 2(a(? ) + 2 X7 (flg ) -> 2 X"fir (a?) + / 2( ,)

Double Replacement

For double replacement reactions two reactants will react by having the cations exchange places with the
anions. The key to this type of reaction, as far as identifying it over the other types, is that it has two
compounds as reactants (or before the arrow).

This type of reaction is more common than any of the others and there are many different types of double
replacement reactions. Some double replacement reactions are more common than others. For example,
precipitation and neutralization reactions are two of the most common double replacement reactions.
Precipitation reactions are ones where two aqueous compound reactants combine to form products where
one of the products is an insoluble solid. A neutralization reaction is one where the two reactant compounds
are an acid and a base and the two products are a salt and water (i.e. acid + base -» salt + water).

General equation for a double displacement reaction: AB + XY — ^ XB + AY

t f

In this reaction, the reactant compounds exchanged partners. A broke up with B and joined Y and X broke up with Y and
joined B.

Example: AgN0 3{aq) + NaCl {aq) -^ AgCl {s) + NaN0 3(aq)
(this is a precipitation reaction because AgCL s \ is formed)

2 NaOH {aq) + H 2 SO A{aq) -» Na 2 S0 4{aq) + 2 H 2 {L)

(this is a neutralization reaction because the acid, H 2 S04, is neutralized by the base, NaOH)

In order to write the products for a double displacement reaction, you must be able to determine the
correct formulas for the new compounds. Let's practice with an example or two.

A common laboratory experiment involves the reaction between lead(II) nitrate and sodium iodide, both
clear solutions. Here is the start of the reaction.

Pb(N0 3 ) 2{aq} + Nal {aq) ->

Now, predict the products from your knowledge of oxidation numbers (or charges).

We know that the cations exchange anions. We now have to look at the charges of each of the cations and
anions to see what the products will be.

We should presume the oxidation number of the lead will remain 2+ and since iodine forms ions with an
oxidation number of 1-, one product will be Pbl 2 . The other product will form between sodium ions whose
oxidation number is 1+ and nitrate ions whose oxidation number is 1-. Therefore, the second product will
be NaN0 3 . Once the products are written in, the equation can be balanced.

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Pb(N0 3 )2(aq) + 2 Nal {aq) -» Pbl 2{s) + 2 NaN0 3{aq) .

The experiment produces a brilliant yellow precipitate. If you have use of a solubility table, it is easy to
determine that the precipitate will be the lead (II) iodide. Lead compounds tend to precipitate and sodium
compounds are always soluble so we would know that the Pbl 2 is the brilliant yellow precipitate.

Look at the reaction between acetic acid and barium hydroxide below.

HC 2 H 3 2{ag) + Ba{OH) 2{aq) -» ?

Predict the products in the same manner as you did in the previous reaction by having the cations exchange
places and writing the correct formulas for the products formed.

HC 2 H 3 2 (aq) + Ba(OH) 2 ( aq ) -> Ba{C 2 H 3 2 ) 2{aq) + H 2 0^ (not balanced)

Therefore the final reaction will be: 2 HC 2 H 3 2 ^ + Ba(OH) 2 ^ aq -j — > Ba(C 2 H 3 2 ) 2 ^ + 2 H 2 0^ (balanced)

This is an acid-base reaction yielding the salt, barium acetate, and water. Notice that HOH and H 2 are
the same.

Sample Question:

(a) Write a double replacement reaction for the reaction between calcium chloride solution and potassium
hydroxide solution to produce potassium chloride solution and a precipitate of calcium hydroxide.

(b) Predict the products for the following reaction: AgN0 3 ( aq ^ + NaCh aq \ -»

(a) CaCl 2{aq) + 2 KOH (aq) -» Ca(OH) 2{s) + 2 KCl {aq)

(b) AgN0 3(aq) + NaCl {aq) -» AgCl {s) + NaN0 3(aq)

Combustion Reactions

A special type of single displacement reaction deserves some attention. These reactions are combustion
reactions, specifically ones that involve fuels such as hydrocarbon compounds. In a particular branch of
chemistry, known as organic chemistry, we study compounds known as hydrocarbons. You might remember
this from previous science classes. In any event, hydrocarbons (hydrogen + carbon) represent the major
components of all organic material including fuels. Combustion reactions usually have the same products,
C0 2 and H 2 0, and one of its reactants is always oxygen. In other words, the only part that changes from
one combustion reaction to the next is the actual hydrocarbon that combusts. The general equation is
given below. Notice the oxygen, carbon dioxide, and water parts of the reaction are listed for you to show
you how these reactants and products remain the same from combustion reaction to combustion reaction.

General equation: Hydrocarbon + 02(g) — * C0 2 ^ + H 2 0^

Also, combustion reactions are all exothermic. Small hydrocarbons (hydrocarbons having 1 to 4 carbon
atoms in them) are gases at room temperature; all of the rest are liquids. Look at the reaction for the
combustion of octane, CsH±s, below. Octane has 8 carbon atoms hence the prefix "oct".

Example: 2 C 8 #i 8{L) + 25 2(g) -> 16 C0 2{g) + 18 H 2 {L)

Now you try one. Write the combustion reaction for the combustion of hexane, Ce^i4(L)-

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Step 1: We know the general equation: Hydrocarbon + 02(g) ~ * ^02(g) + H 2 On\, now put in our hydro-
carbon, hexane, into the general equation.

C&H\A{L) + 02(g) -» C0 2 ( g ) + #20(L)

Step 2: We now have to balance the reaction. Remember we have to start with the carbon atoms, then
the hydrogen and finally oxygen. Do you know why? Carbon is first because it is considered a main
element - so we put a coefficient of 6 in front of the carbon dioxide in the products. Then if we look
at hydrogen and oxygen, hydrogen is in 2 compounds (species: CqHu and H 2 0) and oxygen is in 3
elements/compounds (species: 2 ,C0 2 , and H 2 0). Since hydrogen appears fewer times, we do hydrogen
before oxygen. Therefore, we would put a coefficient of 7 in front of the water in the products.

C6#i4(L) + 02(g) -» 6 C0 2 ( g ) + 7 H 2 0( L )

In order to balance the oxygen, we have to insert a coefficient of -j in front of the oxygen gas in the
reactants.

C6#14(L) + Y°2fe) -* 6 C02(g) + ' #20(L)

Step 3: Remove the fraction. (Multiply the entire reaction by 2)

2 C 6 #i4(L) + 19 2{g) -» 12 C0 2{g) + 14 H 2 (L)

Sample Question: Write the combustion reaction for the combustion of butane, C&H\q.
Solution:

2 C^HiQ^g) + 13 2 (g) -* 8 C0 2 ( g ) + 10 //20(l)

The reactions we have done so far are what are referred to as complete combustion. Complete combustion
reactions occur when there is enough oxygen to burn the entire hydrocarbon. This is why there are only
carbon dioxide and water as products. Have you ever, though, been in a lab, heating a beaker and seen the
black soot appear on the bottom of the beaker. Or seen the black puffs of smoke come out from the exhaust
pipe of a car? If there is not enough oxygen the result is an incomplete combustion reaction with COt g \
and C soot also formed as products. Incomplete combustion reactions are actually quite dangerous because
one of the products in the reaction is not carbon dioxide but carbon monoxide. Carbon monoxide is the
gas that causes people to become sleepy when they breathe it and then eventually, when the concentration
in the blood becomes too high, the person dies.

Sample Question: Identify whether each of the following reactions are complete or incomplete combustions
and then balance each reaction.

(a) C 7 #i 6 ( L ) + 2 (g) -» C0 2 ( g ) + H 2 0( L )

(b) C 3 // 8(g ) + 02( g ) -» C0 2 ( g ) + #20(Z,)

(d) C 5 // 12 (L) + 02(g) -> C 02(g) + #20(L)

(e) C 2 # 6 ( g ) + 2 (g) -» CO^-j + #20(L)

Solution:

(a) Complete; C 7 H 16 ^ + 11 2(g) -» 7 C0 2 ( g ) + 8 # 2 0(l)

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(b) Complete; C 3 H 8{g) + 5 2{g) -> 3 C0 2{g) + 4 H 2 {L)

(d) Complete; C 5 H 12 ( L ) + 8 2 ( g ) -» 5 C0 2 (s) + 6 H 2 0( L )

(e) Incomplete; 2 C 2 H 6 ^ + 5 2 (g) ~* 4 C#(g) + 6 # 2 0(l)

Lesson Summary

• There are five types of chemical reactions (Table 12.3):

Table 12.3:

Reaction Name Reaction Description

Synthesis: two or more reactants make one product.

Decomposition: one type of reactant makes two or more products.

Single replacement: one element species reacts with one compound

species to form products.

Double replacement: two compounds act as reactants.

Combustion: a fuel reactant reacts with oxygen gas.

Review Questions

1. When balancing combustion reactions, did you notice a consistency relating to whether the number
of carbons in the hydrocarbon was odd or even?

2. Distinguish between synthesis and decomposition reactions.

3. When dodecane, C\qH 22 , burns in excess oxygen, the products would be

(a) C0 2 + 2H 2

(b) CO + H 2

(d) c// 4 o 2

4. In the decomposition of antimony trichloride, which of the following products and quantities will be
found?

(a) An + Cl 2

(b) 2 An + 3 Cl 2

(d) 2 Sb + 3 Cl 2

5. Acetylsalicylic acid (Aspirin ), CgHgO^s), is produced by reacting acetic anhydride, C^HqO^I),
with salicylic acid, CjHqO^s). The other product in the reaction is water. Write the balanced
chemical equation.

6. When iron rods are placed in liquid water, a reaction occurs. Hydrogen gas evolves from the container
and iron(III) oxide forms onto the iron rod.

(a) Write a balanced chemical equation for the reaction.

(b) What type of reaction is this?

7. A specific fertilizer is being made at an industrial plant nearby. The fertilizer is called a triple
superphosphate and has a formula Ca{H 2 PO '4)2- It is made by sand and clay that contains phosphate

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and then treating it with a calcium phosphate solution and phosphoric acid. The simplified reaction is
calcium phosphate reacting with the phosphoric acid to yield the superphosphate. Write the balanced
chemical reaction and name the type.

Further Reading / Supplemental Links

Website with lessons, worksheets, and quizzes on various high school chemistry topics.

• Lesson 8-2 is on the Classification of Chemical Reactions, http : //www . f ordhamprep . org/gcurran/
sho/sho/lessons/lesson82 . htm

• http : //library .thinkquest .org/2923/react .html

• http://en.wikipedia.org

Vocabulary

synthesis A synthesis reaction is one in which two or more reactants combine to make ONE type of
product. (A + B ^ C).

decomposition A decomposition reaction is one in which ONE type of reactant breaks down to form
two or more products. (C — » A + B).

single replacement (metal) In a single replacement (metal) reaction, one element replaces the metal
cation of the compound reactant to form products. {A + BC — » AC + B).

single replacement (many metals with acid) In a single replacement (many metals with acid) reac-
tion, one element replaces the hydrogen cation of the compound (which is an acid) reactant to form
products. Example: A + 2 HC — > AC2 + Hi-

single replacement (non-metal) In a single replacement (non-metal) reaction, one element replaces
the non-metal (anion) of the compound reactant to form products. (XY + Z — » XL + Y).

double replacement For double replacement reactions two reactants will react by having the cations
replace the anions. (AB + XY -» AY + XB).

combustion (complete) Combustion is the burning in oxygen, usually a hydrocarbon. ( fuel + O2 — »
CO2 + H2O).

combustion (incomplete) Incomplete combustion is the inefficient burning in oxygen, usually a hydro-
carbon. Inefficient burning means there in not enough oxygen to burn all of the hydrocarbon present,
sometimes carbon (soot) is also a side product of these reactions. ( fuel + O2 — » CO2 + H2O).

hydrocarbons Compounds containing hydrogen and carbon.

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Labs and Demonstrations for Chemical Reactions
Chemical Reactions in Microscale

Investigation and Experimentation Objectives

In this activity, the student will see the various types of chemical reactions and will associate particular
observations with specific reaction types.

Purpose

To write balanced chemical reactions, identifying the type of reaction, and the physical characteristics that
indicate a reaction has taken place.

Background

Within the five main types of chemical reactions studied in the Chemical Reactions chapter, four of these
(synthesis, decomposition, single replacement, and double replacement), have subgroups of reactions that
can be classified. These subgroups are known as precipitation reactions, neutralization reactions, combus-
tion reactions, and the like. Physical changes such as: the formation of a precipitate (hence the precipitation
reaction subclass), change in color, gas formation, change in temperature, tell us that a reaction has taken
place. As well, within each of these reactions, whether an observable physical change has occurred or not,
the Law of Conservation of mass is always maintained.

In this lab you are given seven nitrate solutions with which you are going to react seven sodium and one
ammonium solutions. Planning is everything! Added to this you are only working with 10 drops of solution
in total.

Materials

0.1 mol/L Cu{N0 3 ) 2

0.1 mol/L Na 2 C0 3

0.1 mol/L Pb(N0 3 ) 2

0.1 mol/L Na 2 S0 4

0.1 mol/L Ni{N0 3 ) 2

0.1 mol/L NaCl

0.1 mol/L Co(N0 3 ) 3

0.1 mol/L Nal

0.1 mol/L Fe(N0 3 ) 3

0.1 mol/L Na 2 CrC>4

0.1 mol/L AgN0 3

0.1 mol/L Na 2 Cr 2 7

0.1 mol/L HN0 3

0.1 mol/L NaOH

0.1 mol/L NH^OH

H 2

24-well micro plate

toothpicks

beral pipettes

24-well micro plate (x 4)

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Table 12.4: Data Table

A
B
C
D
E
F
G

Safety

Avoid contact with the solutions. If solutions get on your skin, rinse the area thoroughly with running
water.

Procedure

Part 1: Place 4 micro plates in a grid so that you have at least 8 wells in a line and 7 lines down.

Part 2: In wells Al through to A8, add 5 drops of 0.1 mol/L Cu(N0 3 ) 2 .

Part 3: In wells Bl though B8, add 5 drops of 0.1 mol/L Pb(N0 3 ) 2 .

Part 4: In wells CI though C8, add 5 drops of 0.1 mol/L Ni{N0 3 ) 2 .

Part 5: In wells Dl though D8, add 5 drops of 0.1 mol/L Co(N0 3 ) 3 .

Part 6: In wells El though E8, add 5 drops of 0.1 mol/L Fe(N0 3 ) 3 .

Part 7: In wells Fl though F8, add 5 drops of 0.1 mol/L AgN0 3 .

Part 8: In wells Gl though G8, add 5 drops of 0.1 mol/L HN0 3 .

Part 9: In wells Al through to Gl, add 5 drops of 0.1 mol/L Na 2 C0 3 .

Part 10: In wells A2 though G2, add 5 drops of 0.1 mol/L Na 2 SOi.

Part 11: In wells A3 though G3, add 5 drops of 0.1 mol/L NaCl.

Part 12: In wells A4 though G4, add 5 drops of 0.1 mol/L Nal.

Part 13: In wells A5 though G5, add 5 drops of 0.1 mol/L Na 2 CrO^.

Part 14: In wells A6 though G6, add 5 drops of 0.1 mol/L Na 2 Cr 2 0^.

Part 15: In wells A7 though G7, add 5 drops of 0.1 mol/L NaOH.

Part 16: In wells A8 though G8, add 5 drops of 0.1 mol/L NH^OH.

Part 17: Record all of your observations into your data table.

Part 18: Clean Up. Empty the contents of the micro plate into the sink and rinse the plate and the sink
with plenty of water. Wash your hands and the container thoroughly.

Analysis

1. Which reactions resulted in the formation of a precipitate?

2. Write balanced chemical equations for the reactions found in question 1. Can you determine based
on your observations in this lab what the precipitate is likely to be? If so, indicate that in your
chemical reaction.

3. What other physical changes were observed?

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4. Write the balanced chemical equations for these reactions.

5. What three questions do you have as a result of doing this experiment?

Conclusion

What conclusions did you make as a result of doing this experiment?

Types of Chemical Reactions

Investigation and Experimentation Objectives

In this activity, the student will see the various types of chemical reactions and will associate particular
observations with specific reaction types.

Pre-lab Questions

1. How does a decomposition reaction differ from the other types of chemical reactions?

2. In a combination reaction, what other products form in addition to any new compound?

3. Which two ions are bioaccumulative and should be used in very small quantities?

4. During which steps (give numbers) of the procedure would you expect to produce a gas?

5. What safety precaution applies when heating a test tube?

Introduction

In this experiment, you will learn to differentiate among five general types of chemical reactions. You
will carry out certain representative reactions yourself, while others will be demonstrated by your teacher.
From your observations you will attempt to identify the products of each reaction, and to determine the
type of reaction that has taken place. The types of reactions that you will consider are the following:
synthesis reactions, decomposition reactions, single replacement reactions, double replacement reactions,
and combustion reactions. The majority of common chemical reactions can be classified as belonging to
one of these categories. A brief description of each reaction type is provided below.

(a) Synthesis reactions are reactions in which two or more substances combine to form a single product.
The reactants may be elements or compounds, but the product is always a single compound. An example
of a combination reaction is a reaction of sulfur trioxide and water to form sulfuric acid.

SO m +H 2 (L) ^>H 2 S0 4 (aq)

(b) Decomposition reactions are reactions in which a single substance breaks down into two or more
simpler substances. There is always just a single reactant in a decomposition reaction. An example of a
decomposition reaction is the breakdown of calcium carbonate upon heating.

CaCO^fA + heat — > CaO^ + C0 2 (g)

(c) Single replacement reactions are reactions in which an element within a compound is displaced to
become a separate element. This type of reaction always has two reactants, one of which is always an
element. An example of a single replacement reaction is the reaction of zinc metal with hydrochloric acid.

Zit( s ) + 2 HClfa) -» ZnCI 2 ( aq ) + H 2 ( g )
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(d) Double replacement reactions are reactions in which a positive ion from one ionic compound exchanges
with the positive ion from another ionic compound. These reactions typically occur in aqueous solution
and result in either the formation of a precipitate, the production of a gas, or the formation of a molecular
compound such as water. An example of a double-replacement reaction is the reaction that occurs between
aqueous silver nitrate and aqueous sodium chloride. A precipitate of solid silver chloride is formed in the
reaction.

AgN0 3(aq) + NaCl {aq) -> AgCl {s) + NaN0 2(aq)

(e) Combustion reactions are reactions in which an element or compound reacts rapidly with oxygen gas
to liberate heat and light energy. Commonly, the compounds combining with oxygen in these reactions are
hydrocarbons, compounds containing hydrogen and carbon. The well-known combustible fuels kerosene
and gasoline, for instance, are hydrocarbon mixtures. The complete combustion of a hydrocarbon yields
carbon dioxide and water as well as reaction products. If insufficient oxygen is available, combustion will
not be complete and carbon monoxide and elemental carbon may be obtained as additional products in
the reaction. An example of a combustion reaction is the burning of methane gas to give water (in the
form of steam), carbon dioxide, heat, and light.

CH A{g) + 2 2(g) -» C0 2 ( g) + 2 H 2 (g) + heat + light
Objectives:

• To observe chemical reactions in order to determine the reaction type.

• To write balanced chemical equations for each reaction.

Apparatus and Materials

Iron filings

safety glasses

Copper(II) sulfate pentahydrate

2 x small test tubes

Magnesium, turnings

2 X medium test tubes

0.1 mol/L copper(II) sulfate

lx large test tube

0.2 mol/L lead(II) nitrate

1 x test tube holder

0.2 mol/L potassium iodide

gas burner

3% hydrogen peroxide

ring stand & clamp

5 mol/L hydrochloric acid

dropper pipette

3% sulphuric acid (teacher demo)

crucible tongs

sodium bicarbonate (teacher demo)

electrolysis apparatus (teacher demo)

limewater (teacher demo)

1-holed rubber stopper (teacher demo)

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• toothpicks

• 1 glass tube, 25 cm long bent at 90° in

• matches

Procedure

1. Draw a table similar to the one below to use for collecting your observations.

Table 12.5: Observations

Reaction Observations Reaction Type

Fe and CuSO±
Pb(N0 3 ) 2 and KI
C11SO4 ■ 5H2O and heat
Mg and HCl
H2O2 and heat
Electrolysis of H 2
NaClOi and heat

2. Iron metal and copper(II) sulfate solution. Fill a small test tube halfway with copper(II) sulfate
solution. Add 2 g (about \ of a small test tube) of iron filings to the solution. Observe the reaction after
5 minutes. Record your observations in the observation table. Discard the solid contents of the test tube
into the waste container provided. The liquid portion can be poured down the sink.

3. Lead(II) nitrate and potassium iodide solutions. Put 2 mL of lead(II) nitrate solution in the test
tube. Add 5 to 10 drops of potassium iodide solution. Record your observations. Discard the contents of
the test tube into the waste container and rinse the tube with water.

4. Action of heat on copper(II) sulfate solution. Put two or three pea-sized crystals of copper(II)
sulfate pentahydrate into a large, dry test tube. Fasten a utility clamp to the upper end of the test tube.
Hold the tube by the clamp so that it is almost parallel with the surface of the lab bench. CA UTION:
Do not point the open mouth of the tube at yourself or anyone else. Heat the crystals gently
at the bottom of the tube (where the crystals are located) in a burner flame for approximately 30 seconds
recording your observations. When the test tube has cooled, discard its contents into the waste container
provided.

5. Magnesium metal and hydrochloric acid. Fill one medium-size test tube halfway with 6 mol/L
hydrochloric. CAUTION: Hydrochloric acid is corrosive. Place the test tube in the test tube rack.
Put several pieces of magnesium turnings into the acid solution. If you observe a gas forming, test for
its identity by holding a burning wood splint at the mouth of the test tube. Do not put the splint into
the solution. Record your observations. Decant the liquid portion of the test tube contents into the sink;
discard the solid into the waste container provided.

6. Action of heat on hydrogen peroxide. Add 2 mL of the 3% hydrogen peroxide solution to a medium
test tube. Use a utility clamp to secure the tube to a ring stand. CAUTION: Make sure that the
mouth of the tube is pointed away from you and away from everyone else. Heat the solution
very gently. If you observe a gas forming, test for its identity by inserting a glowing wood splint at the
mouth of the test tube. Do not put the splint into the solution. Record your observations. Rinse the
contents of the test tube into the sink.

TEACHER DEMONSTRATIONS

7. Action of electricity on water (Electrolysis). Water can be broken down to its component

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elements by passing electricity through it. This process is called electrolysis. The apparatus used for
this demonstration will be explained by your teacher. Make your observations of the reaction at several
intervals during a period of 5 - 10 minutes.

8. Action of heat on sodium bicarbonate. Solid sodium bicarbonate will be heated strongly in a
test tube for 2 minutes. The gas that is given off will be tested by exposing it to a burning splint, and
by bubbling it through limewater (a saturated solution of calcium hydroxide, Ca{OH)2). Record your
observations of these tests.

Data Analysis

1. Decide which type of reaction is represented by each reaction observed in this experiment. Record
your answers in your observation table.

2. Write a balanced chemical equation for each chemical reaction observed.

3. Although you did not work with any synthesis reactions in this experiment, can you describe one or
give an example of one that you might have seen before or read about. Write a balanced equation
for this reaction.

Results and Conclusions

1. Describe in your own words the five types of chemical reactions that were discussed in the introduction
to this experiment. Explain how each type of reaction can be identified.

2. List the tests that were used in this experiment to identify gases.

Extension

1. Make a list of the reactions observed in previous experiments. Identify the types of reaction in as
many cases as possible.

Image Sources

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Chapter 13

The Liquid State

13.1 The Properties of Liquids

Lesson Objectives

• The student will explain the basic behavior and characteristics of liquids using the molecular arrange-
ment in liquids.

Introduction

Liquids and solids differ from gases in two important ways. The particles (atoms, molecules, or ions) are
much closer together so the total volume of a liquid or solid is much closer to the sum of the volumes of
the molecules. Also, attractive forces between the particles in liquids and solids are much stronger. The
strength of these forces of attraction between particles is a major reason that the substance is in the liquid
or solid state rather than gaseous.

Liquids Maintain Their Volume But Take the Shape of Their
Container

The arrangement of the molecules in a liquid structure accounts for most of the physical properties of
liquids (see Figure 13.1).

S AS

& W ysft

LIQUID

SOLID

Figure 13.1: The Three Phases of Matter.

Liquids have much less space between molecules and stronger attractive forces than gases. In the liquid
state, attractive forces between molecules are a major factor in the behavior of the liquids. Although liquid
structure has small spaces so the molecules can move past one another, the attraction between particles

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keep them from getting very far apart. Therefore, liquids maintain their own volume but take the shape
of their container. 100 mL of liquid will be 100 mL in any container but because the liquid molecules are
not held in a tightly packed pattern (like solids), the molecules can move past one another and so liquids
will flow to fit the shape of their container

Liquids Have Greater Densities Than Gases

In gases, the distance between the molecules is so great that the size of the molecules themselves become
inconsequential. A gas is considered to be essentially empty space. (If the picture above were realistic
of gas structure, the molecules would be so small you couldn't see them.) If we consider the volumes of
6.02 X 10 23 molecules of oxygen gas (O2) and 6.02 X 10 23 molecules of Freon gas (CF4) both at STP, we
know that the volumes will be the same, 22.4 liters. This is because the sizes of the molecules themselves
are negligible compared to the empty space in a gas. The fact that the Freon molecules are several times
larger than the oxygen molecules makes no difference. In a gas, you are measuring the volume of the empty
space. If we consider the volume of 6.02 x 10 23 molecules of liquid oxygen and 6.02 x 10 23 molecules of
liquid Freon both at STP, we find the volume of the oxygen is about 28 mL and the volume of the Freon
is about 55 mL. In the case of liquids, the volumes of the molecules themselves make a difference. In
liquids, the volume of a group of molecules is related to the volume of the individual molecules and so
equal moles of liquids DO NOT occupy equal volumes under the same conditions. Since liquids have many
more molecules in a smaller volume, they will have much greater densities than gases.

Liquids Are Almost Incompressible

When a substance is compressed, it is not the molecules themselves that are compressed; it is the space
between the molecules that is compressed. Gases have lots of empty space and so are easily compressed.
A pressure of 3.0 atm compresses a gas to one-third its volume at 1.0 atm. Liquids have very little space
between molecules and do not compress easily - a pressure of 3.0 atm will have virtually no effect on
the volume of a liquid. Liquids are used in hydraulic systems because of their ability to flow to fit their
container and their incompressibility.

Liquids Diffuse More Slowly Than Gases

Diffusion in gases (mixing) is nearly instantaneous. If you release a colored gas in a container of non-colored
gas, the color spreads evenly throughout the container in a second or two. In liquids, diffusion is a much
slower process. The smaller spaces to move through and the almost constant collisions cause the molecules
to require much more time to move from one side of the container to the other.

Lesson Summary

• Molecules in the liquid phase have some freedom of movement but their motion is much more re-
stricted than that of gases.

• Liquids are only slightly compressible.

• Liquids diffuse more slowly than gases.

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Review Questions

1. The molar volumes of solid silicon and solid bromine under the same conditions are different and
the molar volumes of liquid silicon and liquid bromine under the same conditions are different, but
the molar volumes of gaseous silicon and gaseous bromine under the same conditions are exactly the
same. Explain.

Vocabulary

incompressible The terms compressibility and incompressibility describe the ability of molecules in a
fluid to be compacted (made more dense).

13.2 Forces of Attraction

Lesson Objectives

• The student will identify liquids whose intermolecular forces of attraction are due to London disper-
sion forces, polar attractions, and hydrogen bonding.

• The student will describe some of the unique properties of water that are due to hydrogen bonding.

• The student will select from comparative compounds, the ones most likely to form hydrogen bonds.

• The student will select from comparative compounds whose intermolecular forces are London disper-
sion forces, the one most likely to have the strongest intermolecular forces.

Introduction

Forces on the molecular level are divided into two categories - the forces inside a molecule holding atoms
together to form the molecule and the forces between molecules holding the molecules in the liquid or solid
state. The forces inside the molecule are called intramolecular forces and the forces between molecules are
called intermolecular forces. You can relate them to bus systems where intracity buses move people around
inside one city and intercity buses move people from one city to another.

Intramolecular Forces

Intramolecular forces refer to the forces inside a molecule that holds the atoms together to form molecules.
Essentially, there is only one force in this category and that is covalent chemical bonds. See below figure.

The only intramolecular forces are covalent chemical bonds. (Source: Richard Parsons. CC-BY-SA)

\ • I

Even though we sometimes write ionic compounds in equations and call the equations "molecular equa-
tions", ionic compounds are not considered to form molecules. Ionic compounds in the solid state from
crystal lattices in which each ion is equally attracted to more than one oppositely charged ion. The only
true molecules are in covalently bonded compounds.

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Intermolecular Forces of Attraction

We will consider three types of intermolecular forces of attraction. These are the forces that hold molecules
in the condensed states (liquid and solid). See below figure.

Intermolecular forces of attraction are the forces between molecules that hold them in the liquid or solid
state. (Source: Richard Parsons. CC-BY-SA)

H *v _

When a substance changes phase as in a solid melting to liquid or a liquid vaporizing to gas, the inter-
molecular forces must be overcome. Therefore, the stronger the intermolecular forces of attraction, the
greater the molecular motion (temperature) that will be required to overcome them. The liquids with the
strongest intermolecular forces of attraction will have the highest boiling points and vice versa.

London Dispersion Forces

The weakest type of intermolecular force is called London dispersion forces. London dispersion forces
occur between all atoms and molecules but they are so weak, they are only considered when there is no
other intermolecular force. For example, London dispersion forces exist between water molecules but water
molecules also have a permanent polar attraction so much stronger than the London dispersion forces that
the London dispersion force is insignificant and not mentioned.

The cause of London dispersion force is not obvious. Although we usually assume the electrons of an atom
are uniformly distributed around the nucleus, this is apparently not true at every instant. As the electrons
move around the nucleus, at a given instant, more electrons may be on one side of the nucleus than the other
side. This momentary nonsymmetrical electron distribution can produce a temporary dipolar arrangement
of charge. This temporary dipole can induce a similar dipole in a neighboring atom and produce a weak,
short-lived attraction.

The cases where London dispersion forces would be considered as the intermolecular force of attraction
would be in the noble gases and non-polar molecules such as H2,N2,Cl2,CH4,CCU,CC>2,S F§ and so forth.
Since non-polar molecules do not have a permanent dipole and no further bonding capacity, their only means
of attracting each other is through London dispersion forces. Some of the substances whose intermolecular
forces of attraction are London dispersion forces are held in the liquid state very weakly and therefore,
would have the lowest melting points of all substances. (See Table 13.1).

Table 13.1: Boiling Points of Some London Dispersion Forces Liquids

Substance Boiling Point (°C)

Helium, He -269.7

Neon, Ne -248.6

Argon, Ar -189.4

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Table 13.1: (continued)

Substance Boiling Point (°C)

Krypton, Kr -157.3

Xenon, Xe -111.9

Hydrogen, H2 -253

Oxygen, 2 -182

Methane, CH A -161

Carbon Dioxide, CO2 -78

The temporary dipoles that cause London dispersion forces are affected by the molar mass of the particle.
The greater the molar mass of the particle, the greater is the force of attraction caused by London dispersion
forces. The molar masses of H2, N2, and O2 are 2, 28, and 32 g/mol, respectively, and their boiling points
increase in similar fashion; H2 is -253°C, N2 is — 196°C, and O2 is -183°C. When the molar masses of
London dispersion force liquids become high enough, the substances will be liquid even at room temperature
and above. Carbon tetrachloride, molar mass 154 g/mol, and bromine, molar mass 160 g/mol, boil at
+77°C and +59°C, respectively. Many long carbon chain, non-polar substances such as gasoline and oil
remain liquids at common temperatures.

Polar Attractions

When covalent bonds form between identical atoms such as in H2, N2, O2, and so on, the electrons shared
in the bonds are shared exactly equally. The two atoms have the same electronegativity and therefore, the
same pull on the shared electrons. (See Figure 13.2).

Center of Negative Charge Center of Positive Charge

Figure 13.2: The centers of positive and negative charge.

The center of negative charge for the entire molecule will be in the exact center of the molecule. This will
coincide with the center of positive charge for the molecule. When the center of negative charge and the
center of positive charge coincide, there is no charge separation and no dipole.

If the two atoms sharing the bonding pair of electrons are not of the same element, the atom with the greater
electronegativity will pull the shared electrons closer to it. Because of the resulting uneven distribution of
electrons, the center of negative charge will not coincide with the center of positive charge and a dipole
is created on the molecule. When the centers of positive and negative charge do not coincide, a charge
separation exists and a dipole is present.

The end of the molecule with the more electronegative atom will have a partial negative charge and the
end of the molecule with the more electropositive atom will have a slight positive charge. The symbols 8+
and 6- are used because these are not full 1+ and 1- charges. See below figure.

The polar molecule. (Source: Richard Parsons. CC-BY-SA)

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This polarity is much less of a charge separation than in an ionic bond. These charges are only fractions
of full 1+ and 1- charges. How much polarity a bond will experience depends on the difference in the
electronegativities of the atoms.

In the case of a symmetrical molecule with polar bonds, the symmetry of the electron displacements of
two or more electron pairs will keep the center of negative charge in the center of the molecule coincident
with the center of positive charge and so no dipole will occur. See below figure.

When the electron shift in polar bonds is symmetrical over the entire molecule, there will be no dipole.
(Source: Richard Parsons. CC-BY-SA)

:0 = C =

■ ■

For example, in the CO2 molecule, both carbon-oxygen bonds are polar, but the shift of bonding electrons
toward the oxygen is the same on both sides of the carbon atom and the center of negative charge remains
in the center.

For molecules that have a permanent dipole, the attraction between oppositely charged ends of adjacent
molecules are the dominant intermolecular force of attraction.

Figure 13.3: Polar attractions in a solid hold the molecules together.

Figure (13.3) represents a polar solid, a polar liquid would look similar except the molecules would be
less organized. On the average, these polar attractions are stronger than London dispersion forces so
polar molecules, in general, have higher boiling points than London dispersion liquids. There is significant
overlap, however, between the boiling points of the stronger London dispersion molecules and the weaker
polar molecules.

The organization of a substance composed of polar molecules depends on the competition between the
strength of the polar attractions and the molecular motion of the molecules. At higher temperatures, the
molecular motion of the molecules is strong enough to disrupt the polar attractions but at low tempera-
tures, the molecular motion is reduced and the polar attractions can hold the molecules in a structured
arrangement. In liquid and gaseous forms, the molecules can also turn freely. If a pair of polar molecules
in liquid form are aligned positive end to positive end, it is no problem for one of the molecules to spin
around and align its negative end to the positive end of the other molecule. (See Figure 13.4).

This turning of polar molecules toward an attractive force can be seen in a macroscopic situation as well.

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non-polar liquid

water (polar liquid)

charged red

Figure 13.4: A stream of polar liquid will bend toward a charged rod.

If we allow a very thin stream of water to run from a faucet and we bring a charged object (rubber comb
run through hair, balloon rubbed on wool sweater, etc) near the stream of liquid, the stream will bend
its path toward the charged object. It doesn't make any difference if the charged object is positively or
negatively charged because the water molecules in the stream will turn their oppositely charged end toward
the charged object. In the sketch at right, the path of a non-polar liquid is not deflected by the charged
rod but the path of the water stream is deflected by the charged rod.

Hydrogen Bonds

There are several polar molecules whose polar bonds are so strong they merit separate attention. These are
the polar attractions that occur in molecules where hydrogen is bonded to nitrogen, oxygen, or fluorine.
The polar attractions in these molecules are nearly 10 times as strong as regular polar attractions. These
extra strong polar attractions that occur with the H — N, H — O, and H — F bonds are called hydrogen
bonds which distinguishes them from regular polar attractions but keep in mind that they are, in fact,
polar attractions, albeit very strong ones.

There is more than one explanation for why these three combinations form hydrogen bonds. There are
only ten elements that have greater electronegativity than hydrogen and only four that have a significantly
greater electronegativity than hydrogen. Three of the elements that have significantly greater electroneg-
ativity than hydrogen are nitrogen, oxygen, and fluorine - the three elements that form hydrogen bonds
in compounds with hydrogen. When hydrogen bonds with atoms whose electronegativities are less than
or equal to the electronegativity of hydrogen, the other atom cannot pull the shared electrons away from
hydrogen. (See Figure 13.5).

Therefore, the hydrogen nucleus (a single proton) is shielded by electrons in its electron cloud. When
hydrogen chemically bonds with nitrogen, oxygen, or fluorine, the very high electronegativities of these
atoms CAN pull the electrons far away from the hydrogen atom, thus removing the shielding electrons
from the proton nucleus of hydrogen. When the polar attractions between HF molecules or other hydrogen
bonding compounds set up, the negative end of a molecule can get very close to the proton on the positive
end of another molecule because there are no electrons for shielding. The closeness of the charges causes
the extra strong polar attractions in these compounds. The characteristics of a liquid that forms hydrogen
bonds are significantly different from similar compounds that do not form hydrogen bonds. In the case of
water, this is very important to life on earth.

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378

In HBr. the bonding electrons remain
partially around the hydrogen and
shield its nucleus

In HF. the bonding electrons are
pulled away leaving an unshielded
nucleus.

Figure 13.5: With highly electronegative non-metals, the electron shift away from the hydrogen atom leaves
an unshielded proton.

A homologous series of compounds are compounds of the elements of a family, each bonded to the same
other element. For example, family 4A in the periodic table consists of carbon, silicon, germanium, and
tin. If each of these is bonded to hydrogen, it would produce a homologous series, C//4, SiH^, GeH^, and
SnH^. If we graph the boiling points of this homologous series, we would get the graph sketched in Figure
13.6. The large majority of graphs of the boiling points of homologous series would look like this in that
the boiling points increase as molar mass increases. If we graph the boiling points of the homologous series
of family 6A combined with hydrogen, we get a quite different graph.

■200

100

B.P. in C

60 120

Molar Mass

100

60 120

Meier Mass

Figure 13.6: Boiling point comparisons: family 4A hydrogen compounds to family 6A hydrogen compounds.

The higher molar mass compounds in the series follow the normal pattern of decreasing boiling points as
the molar mass decreases. But, when we get to water, suddenly the boiling point is way out of line - it
is more than 150 degrees too high. The explanation for why liquid water is held together far more tightly
than expected - hydrogen bonds. Graphs of the boiling points of the homologous series of hydrogen with
5A family members and hydrogen with 7A family members would be similar. The boiling point of NH3
and HF are greatly different from what would be expected. (See Figure 13.7).

The fact that water forms hydrogen bonds has effects so large that it is impossible in this text to delineate
them all. We will consider a few but there are many more not in this list.

1. The normal boiling point of water is 100°C If water did not form hydrogen bonds and instead had
a regular polar attraction between molecules, its boiling point would be somewhere around — 60°C.
The average temperature of the surface of the earth is 15°C If water did not form hydrogen bonds,
the oceans and lakes would vaporize and earth would not be a water planet. Therefore, it would
likely not be a planet with life on it.

2. If you gathered 100 substances of all types (ionic, metals, regular polar substances, London dispersion

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Water with no significant hydrogen bonding. Water with significant hydrogen bonding.

Figure 13.7: Significant hydrogen bonding causes water molecules to line up end-to-end.

substances) and water and you arranged to have each of these substances in both solid and liquid form
and you dropped pieces of solid into the corresponding liquid, you would find exactly one compound
whose solid floated in the liquid. For all the other compounds, the pieces of solid would sink to
the bottom of the liquid. Only water would have the solid floating on the liquid. The reason for
this is that as you cool the liquid of almost all substances to freeze them, the substance contracts
as it cools - the molecules move around slower and intermolecular forces pull the molecules closer
together. As a result, the solids are denser than the liquids and the solids will sink in the liquids.
Water, of course, is the exception. When water cools to its freezing point and solidifies, it expands.
When water molecules are at or above 4°C, the molecular motion is sufficient to keep the hydrogen
bonds from locking water molecules into an end-to-end molecular complex with large holes in the
structure. When water is cooled below 4°C, the molecular motion is inadequate to break up this
complex structure - so water molecules begin forming end-to-end chains and the water expands
because of the holes in the structure. Therefore, solid water is less dense than liquid water and ice
floats on water. You have seen ice cubes floating in water a thousand times and you have probably
never seen any other substance where you see the solid interacting with the liquid, so the ice floating
looks normal to you. If you were familiar with the behavior of all other substances in this situation
and then you saw someone place ice cubes in water for the first time, you would probably think they
had just done a magic trick. One of the effects of the fact that ice floats on water is that when cold
weather comes to areas in the northern and southern parts of the earth, the cold air contacts the
water on the surface of lakes and freezes that surface water. If water were like other substances and
the solid sank to the bottom, then the cold air would freeze the new surface and it would sink and
so on until the entire lake would be frozen top to bottom. No water dwelling animals would survive
such an occurrence. But, actually, when the cold air freezes the surface of a lake, the ice floats, stays
on top and insulates the rest of the rest of the water from the cold air. Only the surface freezes and
the animals that live in the water survive the winter.

3. One of the factors in the weathering of rocks in geographical areas that have winter is that rain water
enters rocks through cracks and then freezes and expands, fracturing the rock.

4. Some biologically active molecules such as DNA require a particular shape for their function. If you
think of a long-chain molecule as something similar to a string, how can such a molecule have and
hold a shape. At points along its length, the molecule can be linked to itself with different types of
attractions - one of which are hydrogen bonds. See below figure.

5. Hydrogen bonds form links to maintain shapes of molecules. (Source: Richard Parsons. CC-BY-SA)

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7. As long as water is above 4°C, when it is cooled, like other substances, it contracts and becomes
denser. When water passes through 4°C, significant hydrogen bonding begins to form and the water
expands and becomes less dense. The maximum density for water is at 4°C. Many animals that
live in water and require oxygen, use oxygen that is dissolved in the water. Water in lakes becomes
oxygenated (has dissolved oxygen) by the action of wind and waves at the surface. For deep lakes,
diffusion is inadequate to move the oxygenated water to the bottom of the lake. Once the oxygen
was used up in the water at the bottom of a lake, oxygen using animals would be unable to live there.
For lakes in northern climates, the surface passes through the temperature 4°C twice per year, once
as it cools in the fall and once as it warms in the spring. During these two times, the water at the
surface would sink to the bottom because it is denser and the sinking water would be oxygenated.
These periods are called spring and fall "turnovers." The turnovers provide oxygenated water at the
bottom of the lakes.

The special chemical properties of water are explored, along with the need for its protection and conserva-
tion. Water (http : //www . learner . org/vod/vod_window . html?pid=804)

Ionic Liquids

Ionic compounds may also exist in liquid form. The intermolecular forces of attraction in ionic liquids
would, of course, be the electrostatic attraction between oppositely charged ions. The charges on ions are
complete charges of 1+, 2+, 1-, 2- and so on. The attractions due to these charges are much greater than
those of polar molecules, even the especially strong polar attractions of hydrogen bonds, and so the boiling
points of ionic liquids would be much higher than the molecular liquids discussed previously. Sodium
chloride, for example, boils at 1465° C.

Metallic Liquids

Metals may also exist in the liquid state. The bonding and therefore, the intermolecular forces of attraction
for metals will be covered in the next chapter.

Lesson Summary

• Molecules are held together in the liquid phase by intermolecular forces of attractions.

• London dispersion forces are a very weak intermolecular force of attraction caused by a temporary
electrostatic attraction between the electrons of one molecule or atom and the nucleus of another.

• Polar attractions are an intermolecular force of attraction caused by the electrostatic attraction
between permanent dipoles that exists on polar molecules.

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• Hydrogen bonds are an exceptionally strong type of polar attraction that occurs between molecules
that have H — F bonds, H — O bonds, or H — N bonds.

• Ionic substances also exist in liquid form; the attractions between the ions are the attractions between
ionic charges and are much stronger than polar attractions or hydrogen bonds.

Review Questions

1. Which of the following molecules would you expect to have the higher melting point? Why?

(a) CH A or H 2 S

(b) H 2 or H 2 S

(d) Nal or NH 3

(e) S F4 or C//4

2. The structures of vitamins C and E are shown above. Which of the following statements is correct?

(a) Vitamin E has more opportunities for hydrogen bonds than vitamin C.

(b) The melting point of vitamin E is likely to be higher than that of vitamin C.

(d) Vitamin C should have a higher solubility in water than vitamin E.

(e) Vitamin C would be described as a "fat-soluble" vitamin.

Vocabulary

hydrogen bond The exceptionally strong polar attraction between a hydrogen atom in one molecule
and a highly electronegative atom (N, O, F) in another molecule.

London dispersion forces Electrostatic attractions of molecule or atoms for nearby atoms or molecules
caused by the temporary unsymmetrical distribution of electrons in electron clouds.

13.3 Vapor Pressure

Lesson Objectives

• The student will be able to describe the processes of evaporation and condensation.

• The student will be able to describe vapor pressure equilibrium.

Introduction

The phase of a substance is essentially the result of two forces acting on the molecules. The molecules of
a substance are pulled together by intermolecular forces of attraction that were discussed in the previous

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section. Some of these intermolecular forces are weak and some are strong. The molecules of a substance
also have kinetic energy so that the molecules are in constant random motion and are in almost constant
collisions with each other. The motion and collisions of molecules push them away from each other.
Without intermolecular forces of attraction, the molecules of all substances would move away from each
other and there would be no condensed phases (liquids and solids).

If the forces caused by molecular motion are much greater than and dominate the intermolecular forces of
attraction, the molecules will separate and the substance will be in the gaseous state. If the intermolecular
forces of attraction are dominantly stronger than the molecular motion, the molecules will be pulled into
a closely packed pattern and the substance will be in the solid state. If there is some balance between
molecular motion and intermolecular forces of attraction, the substance will be in the liquid state.

When substances are heated or cooled, their average kinetic energy increases or decreases, their molecular
motion increases or decreases, and the substance may change phase. A substance in the solid phase
(intermolecular forces dominate) can be heated until the molecular motion balances the intermolecular
forces and the solid will melt to liquid. The liquid may be heated until the molecular motion completely
overcomes the intermolecular forces and the liquid will vaporize to the gaseous state.

When you open a jar of perfume, your nose detects the substance almost immediately. You can see that
the substance is in liquid form and it is not boiling, yet some of that material obviously entered the gaseous
state and reached your nose.

Evaporation

The temperature of a beaker of water is a measure of the average kinetic energy of the molecules in the
beaker. That does not mean that all the molecules in the beaker have exactly the same kinetic energy.
Most of the molecules will be within a few degrees of the average but a few molecules may be considerably
hotter or colder than the average. The kinetic energy of the molecules in the breaker will have a distribution
curve similar to a standard distribution curve for most naturally occurring phenomena. (See Figure 13.8).

Number of
Occurrences

Vorioble

A standard distribution curve like this could represent the length of
grasshopper legs in a random sample of grasshoppers or the
distribution of 1Q for a random sample of humans.

Figure 13.8: Standard Distribution Curve

Naturally occurring phenomena usually have most of the instances near the average and the number of
instances become less as the value gets farther from the average.

In the case of a beaker of water, some of the molecules will have an average temperature below the boiling
point and some of the molecules will have a temperature above the boiling point (see Figure 13.8). The
dashed yellow line is the average temperature of the molecules and would be the temperature shown on a

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y*X^

Number of

Molecules

/ \

B.P.

-, ^> ■ ■ I —

Temperature

Figure 13.9: The distribution of temperature among the molecules of a beaker of water.

thermometer inserted into the liquid. The red line represents the boiling point of water (100°C at 1.00 atm
pressure) and the area under the curve to the right of the red line represents the molecules of water that
are above the boiling point. In order for a molecule of liquid that is above the boiling temperature to
escape from the liquid, it must either be on the surface or it must be adjacent to many other molecules
that are above the boiling point so that the molecules can form a bubble and rise to the surface. See below
figure.

Water boils only when a sufficient number of adjacent molecules are above the boiling point and can form
bubbles of gaseous water. (Source: Richard Parsons. CC-BY-SA)

The only circumstance when there are enough molecules above the boiling point to form bubbles is when
the average temperature is at the boiling point. For single molecules above the boiling point, they must
wait until their random motion gets them to the surface and then the molecule can leave the liquid and
enter into the gaseous phase. This process of molecules escaping to the gaseous phase from the surface of
a liquid when the average temperature of the liquid is below the boiling point is called evaporation.

The process of phase change is a little more complicated than just having the molecules reach the boiling
point. Objects that attract each other and are separated have potential energy due to that attraction
and separation. An object held above the earth is attracted to the earth by earth's gravity and has
potential energy due to the attraction and separation. The amount of potential energy can be calculated
by multiplying the force of attraction times the distance of separation. Two oppositely charged objects
that are separated have potential energy. Objects with opposite magnetic poles that are separated have
potential energy. A stretched rubber band has potential energy. Gaseous molecules that have a force of
attraction between them but are separated have potential energy. Molecules in the liquid state and the
same molecules in the gaseous state at the same temperature DO NOT have the same total energy. If they
are at the same temperature, they have the same kinetic energy but the gaseous molecules have potential
energy that the liquid molecules do not. Molecules in the liquid state that are hot enough to exist in the
gaseous state must absorb energy from their surroundings to provide that potential energy as they change
phase. This potential energy is called the heat of vaporization and it is absorbed by evaporating or

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vaporizing molecules from the kinetic energy of the liquid.

You are at least somewhat familiar with evaporation. You know that if you leave a saucer of water sitting
out on the countertop, the water will slowly disappear - and yet, at no time is the temperature of the
water ever at the boiling point. The water in an open container continues to evaporate until it is all in the
vapor state. When molecules of a liquid are evaporating, it is clear that it is the hottest molecules that are
evaporating. It might seem that once the molecules whose temperature was above the boiling point are
gone, no more evaporation would occur. Here's the reason evaporation continues. The temperature of the
liquid is the average temperature of all the molecules. When the hottest molecules evaporate, the average
temperature of those molecules left behind is lower and the molecules left behind have also contributed to
the heat of vaporization to the evaporating molecules. The process of evaporation causes the remaining
liquid to cool significantly. Heat flows from warmer objects to colder objects and so when the liquid cools
due to evaporation, the surroundings will give heat to the liquid thus raising its temperature back up equal
to the surroundings thus producing more hot molecules. This process can continue in an open container
until the liquid is all evaporated.

Many years ago, when people lived without electricity, they figured out that if they placed the butter dish
on the dinner table in a shallow container of water, evaporation would cool the water and therefore the
butter dish, enough to keep the butter from melting to a liquid. They also knew that if you put a container
of water or milk in a fabric sack, soaked the sack in water, and swung the sack around in the air, the
evaporation of the water from the sack would cool it and cool the container of milk or water to make it
nicer to drink. Many hikers today use fabric canteen holders that they soak in water while hiking so that
the water in the canteen will be cooler when they drink it.

The rate of evaporation is related to the strength of the intermolecular forces of attraction, to the surface
area of the liquid, and to the temperature of the liquid. As the temperature of liquids get closer to the
boiling point, more of the molecules have temperatures above the boiling point and so evaporation is faster.
Substances with weak intermolecular forces of attraction evaporate more quickly than those with strong
intermolecular forces of attraction. Substances that evaporate readily are called volatile and those that
hardly evaporate at all are called non- volatile.

Condensation

Liquids will usually evaporate to dryness in an open container. What happens, however, if the container is
closed? When a lid is put on the container, the molecules that have evaporated are now kept in the space
above the liquid. This makes it possible for a gaseous molecule to collide with another molecule or a wall
and condense (the gas to liquid phase change) back to liquid. Molecules at the boiling point can exist in
either liquid phase or gaseous phase - the only difference between them is the amount of potential energy
they hold. See below figure.

Evaporation and condensation both occur in a closed container. (Source: Richard Parsons. CC-BY-SA)

For a liquid molecule with adequate temperature to exist in the gaseous phase, it is necessary for the

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molecule to gain the heat of vaporization. It does this by collision with adjacent molecules. For a gaseous
molecule to return to the liquid phase, it must give up the same amount of potential energy that it gained.
That amount of potential energy is called the heat of vaporization when it is being gained and it is
called the heat of condensation when it is being lost - but it is exactly the same amount of energy.

As more and more molecules evaporate in a closed container, the partial pressure of the gas in the space
above the liquid increases. The rate at which molecules evaporate is determined by the temperature, the
surface area, and what substance is involved. Once the substance, the surface area, and the temperature
are established, the rate of evaporation will be constant. The rate at which the gas condenses is determined
by the partial pressure of the gas, the surface area, and what substance is involved. Once the substance
and surface area is established, the rate of condensation will only vary depending on the partial pressure of
the gas. As the partial pressure of the gas in the space above the liquid increases, the rate of condensation
will increase.

In the section on evaporation, it was pointed out that as a liquid evaporates, the remaining liquid cools
because the hottest molecules are leaving so the average decreases and the heat of vaporization is being
absorbed from the remaining molecules. For similar reasons, when a gas is undergoing condensation, the
temperature of the remaining gas increases because the coolest molecules are condensing, thus raising the
average of those left behind, and the condensing molecules must give up the heat of condensation.

Vapor Pressure Equilibrium

You can follow the progress of the two activities (evaporation and condensation) in a thought experiment.
Suppose we place some liquid water in an Erlenmeyer flask and seal it. No water has evaporated yet so
the partial pressure of water vapor in the space above the liquid is zero. As a result, there will be no
condensation. As the water evaporates (at a constant rate since the temperature and surface area are
constant), the partial pressure of the water vapor increases. Now that some vapor exists, condensation
begins. Since the partial pressure of the water vapor is low, the rate of condensation will be low. Over
time, more and more water evaporates and the partial pressure of the water vapor increases. Since the
partial pressure increases, the rate of condensation increases. Eventually, the rate of condensation will
become high enough that it is equal to the rate of evaporation. Once this happens, the rate of molecules
of water going into the vapor phase and the number of molecules condensing back to liquid are exactly
the same and so the partial pressure no longer increases. When the partial pressure of the water vapor
becomes constant, the rate of condensation is constant and is exactly EQUAL to the rate of evaporation.
A condition called vapor pressure equilibrium has been established. As time goes on from this point,
the amount of liquid cannot change, the amount of gas cannot change; neither the rate of evaporation
nor the rate of condensation can change. Everything remains exactly the same, BUT the two activities
continue. Evaporation continues and condensation continues at exactly the same rate. Each different liquid
at each temperature will have an exact partial pressure of vapor that will be present when vapor pressure
equilibrium is established. The pressure of the vapor in the space above the liquid is called the vapor
pressure of that liquid at that temperature. (See Table 13.2).

Table 13.2: Vapor Pressure of Water at Various Temperatures

Temperature in °C Vapor Pressure in Torr

4.6

10 9.2

20 17.5

30 31.8

40 55.3

50 92.5

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Table 13.2: (continued)

Temperature in °C

Vapor Pressure in Torr

60
70
80
90
100

149.4
233.7
355.1
525.8
760.0

Volatile liquids would have higher vapor pressures than water at the same temperature and non-volatile
liquids would have lower vapor pressures at the same temperature. The amount of volume for the space
above the liquid makes no difference. The partial pressures of the gases will reach the equilibrium value
- if the space is small, it will take little gas to produce the pressure and if the space is large, it will take
much more gas to produce the pressure. As long as you introduce enough liquid into the container so that
vapor pressure equilibrium will be reached, then the precise vapor pressure will be attained.

You might have noticed a subtle switch in vocabulary sometimes referring to the substance in the gaseous
state as a gas and sometimes as a vapor. Chemists have agreed that a substance in the gaseous phase
at temperatures above the boiling point of its liquid should be called a gas and if the temperature of the
substance is below the boiling point, it should be called a vapor.

You should also note that the equilibrium vapor pressure of a liquid is the same regardless of whether or
not another gas is present in the space above the liquid. If the space above liquid water contains air at
760 torr, and the liquid water evaporates until its equilibrium vapor pressure (25 torr) is reached, then the
total pressure in the space above the liquid will be 785 torr. The presence of the air in no way affects the
vapor pressure.

Vapor Pressure Correction

When gaseous substances are produced from chemical reactions and collected in the laboratory, they are
usually collected over water. The "collection over water" technique is inexpensive and allows gaseous
substances to be collected without having air mixed in. The process involves filling a collecting jar with
water and inverting the jar in a pan of water without letting any water out or air in.

Figure 13.10: Gas Collection Over Water.

In Figure 13.10, the picture on the far left represents the collecting jar full of water and inverted in a pan
of water. A tube runs from the reaction vessel where the gas is produced and is tucked under the edge of
the collecting jar. As the gas is produced and comes out the end of the tube, it bubbles up through the
water and pushes the water out of the jar. When the water in the collecting jar and the pan are exactly
level, as in the picture at the far right, the pressure inside the collecting jar and the atmospheric pressure
in the lab are equal. Using the pressure in the lab and the temperature in the lab and the volume of the
jar to the water level, you can calculate how much gas you produced. (Plug P, V, T, and R into PV = nRT,

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and solve for n.) It turns out, however, that you must make a correction before you plug in the pressure
value. Since the collecting jar is a closed container and it has liquid water in the bottom of it, then it
will contain the vapor pressure of water at this temperature. So the outside pressure in the lab tells you
the pressure inside the collecting jar but it doesn't tell you how much of that pressure is due to the gas
collected and how much is due to water vapor. You must get a table of the vapor pressure of water at each
temperature and look up the vapor pressure of water at the temperature of your lab and then subtract
that pressure from the total pressure in the collecting jar. The result will be the actual pressure of the gas
collected.

Example

Some hydrogen gas was collected over water in the lab on a day that the atmospheric pressure was 755 torr
and the lab temperature was 20°C Hydrogen gas was collected in the collecting jar until the water levels
inside and outside the jar was equal. What was the partial pressure of the hydrogen in the collecting jar?

Solution

The total pressure in the collecting jar is 755 torr and is equal to the sum of the partial pressure of hydrogen
in the jar and the vapor pressure of water at 20°C From the table, the vapor pressure of water at 20°C is
17.5 torr.

Partial pressure of H2 = 755 torr - 17.5 torr = 737 torr

Lesson Summary

• Molecules of liquid may evaporate from the surface of a liquid.

• When molecules of a liquid evaporate, the remaining liquid cools.

• Gas molecules in contact with their liquid may condense to liquid form.

• If a liquid is placed in a closed container, eventually vapor pressure equilibrium will be reached.

Review Questions

1. A flask half-filled with water is sealed with a stopper. The space above the water contains hydrogen
gas and water vapor in vapor pressure equilibrium with the liquid water. The total pressure of the
two gases is 780. mm of Hg at 20. °C. The vapor pressure of water at 20. °C is 19 mm of Hg. What is
the partial pressure of the hydrogen gas in the flask?

2. Describe all the reasons that the remaining liquid cools as evaporation occurs.

3. Describe all the reasons that the remaining gas gets hotter as condensation occurs.

4. The apparatus above can be used to determine the vapor pressure of benzene. With a vacuum in the
top of the tube, the mercury rises to the height shown. When a small amount of liquid benzene is
injected into the space at the top of the tube, it floats on the mercury. The benzene will evaporate
and eventually reach vapor pressure equilibrium. The mercury in the tube will pushed down further
by the pressure of the benzene vapor in the tube. Neglecting the effect of the liquid benzene, what
would be the calculated vapor pressure of benzene?

5. Water vapor and hydrogen gas are sealed in a cylinder fitted with a piston at 60°C The partial
pressure of the hydrogen gas is 0.35 atm and the vapor pressure of the water is 0.20 atm at this
temperature. The total pressure in the cylinder is 0.55 atm. If the piston is pushed down until the
volume is half the original volume, what will be the pressure in the cylinder?

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Vocabulary

condensation The process whereby a gas or vapor is changed to a liquid.

equilibrium vapor pressure The pressure that is exerted, at a given temperature, by the vapor of a
solid or liquid in equilibrium with the vapor.

evaporation The escape of molecules from a liquid into the gaseous state at a temperature below the
boiling point.

heat of condensation The quantity of heat released when a unit mass of a vapor condenses to liquid
at constant temperature.

heat of vaporization The quantity of heat required to vaporize a unit mass of liquid at constant tem-
perature.

vapor The gaseous phase of a substance that exists even though the temperature is below the boiling
point of the substance.

13.4 Boiling Point

Lesson Objectives

• The student will know the relationship between boiling point, vapor pressure, and ambient pressure.

• Given a vapor pressure table for water, and the ambient pressure, the student will be able to determine
the boiling point of water for those conditions.

Introduction

If you want hard-boiled eggs at home, you can probably put the eggs in boiling water for about eight
minutes to accomplish it. If you go camping in the Rocky Mountains at an altitude of 10,000 feet, you
will find that an egg placed in boiling water for eight minutes is not hard boiled. In fact, even after twelve
minutes in boiling water, the egg may still be a little too runny for your tastes.

Normal Boiling Point

Imagine you are boiling water in a place where the atmospheric pressure is 1.00 atm. In the boiling water,
a large bubble forms near the surface of the liquid water and remaining at the same size rises to the top of
the water and the gas escapes into the air. If the pressure of the gas inside that bubble had been less than
1.00 atm, the outside pressure of the atmosphere would have crushed the bubble and it would not have
existed. If the pressure of the gas inside that bubble had been greater than 1.00 atm, the bubble would
have expanded to a larger size instead of remaining at the same size. The fact that the bubble remained at
the same size indicates that the gas pressure inside that bubble was the same as the atmospheric pressure.

When you are heating water in an effort to boil it, gas bubbles cannot form until the water can produce
a vapor pressure equal to the surrounding air pressure. The hotter the water gets, the higher its vapor
pressure becomes but only when that vapor pressure equals the surrounding atmospheric pressure can the

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water form bubbles in the process we call boiling. A liquid cannot boil until its vapor pressure is equal to
the pressure on the surface of the liquid. The actual definition of boiling point is that temperature at
which the vapor pressure of the liquid equals the surrounding pressure.

If you are measuring boiling points at the normal sea level atmospheric pressure of 1.00 atm, a liquid more
volatile than water such as chloroform will boil at 61.3°C. This is because the vapor pressure of chloroform
is 1.00 atm at 61.3°C. The vapor pressure of ethanol reaches 1.00 atm at a temperature of 78.4°C and
therefore, that is the normal boiling point of ethanol.

Boiling Points Change with Changes in Pressure

Since liquids boil when their vapor pressure becomes equal to surrounding pressure, then if the surrounding
pressure is lower, liquids will boil at lower temperatures. At higher altitudes, atmospheric pressure is lower.
In cities whose altitude is around 5,000 feet, water boils at 95°C instead of at 100°C and at 10,000 feet,
water boils around 90° C. The water boils in normal fashion but its temperature is lower and therefore,
cooking in boiling water takes a longer time. In situations where boiling is used to purify water or sterilize
equipment, the lower temperature of boiling water requires concern.

If a container of water is placed in a bell jar and a vacuum pump attached so that the air pressure around
the water can be greatly reduced, water may be made to boil at very low temperatures. (See Figure
13.11).

To Vacuum Pump

Water Temperature = 20 C

Figure 13.11: A beaker of water under a bell jar with lowered pressure.

At room temperature, 20°C, the vapor pressure of water is 17.5 mm of Hg so if the pressure in the bell
jar is reduced to 17.5 mm of Hg, water will boil at 20° C. The appearance of the boiling water is the same
as it is at 100°C, with steam coming off and so on, but the water can be removed from the bell jar and
poured on your hand and the temperature is only 20° C. When you look up the boiling point of a liquid,
the reference will be to the normal boiling point which means the boiling point when the surrounding
pressure is 1.00 atm.

If the surrounding pressure is less than 1.00 atm, the boiling points of liquids will be lower. Conversely, if
the surrounding pressure is greater than 1.00 atm, the boiling points of liquids will be higher. It's fairly
unusual to find atmospheric pressures greater than 1.00 atm except during storms, but it's easy enough to
raise the surrounding pressure in a laboratory situation. If we use a strong container with a lid that screws
on very tight, we can boil water in the container and as water vaporizes and the temperature of both the
air and the water vapor increase, the gas pressure in the container will increase. As the pressure in the
container increases, the boiling point of the water increases. The vapor pressure of liquid water at 120°C
is 2.0 atm. Therefore, if we can raise the pressure inside a sealed container to 2.0 atm, water will not boil
in the container until its temperature is 120°C This is the concept that is used in pressure cookers and
rice cookers. The cooking pot has a lid that can be sealed tightly and a valve in the lid that will open
slightly when the pressure inside the container reaches 2.0 atm. Water and whatever food you wish to

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cook is placed inside the pressure cooker and it is set on the stove. The pressure and therefore the boiling
point of water increases inside the container until the pressure reaches 2.0 atm. If the pressure goes beyond
2.0 atm, the little valve opens and lets out some gas so that the pressure remains at 2.0 atm. The valve
can be opened and closed any number of times to keep the inside pressure at 2.0 atm. The temperature of
the boiling water inside will be 120°C under these conditions and the food will cook in as little as one-third
the normal time.

Lesson Summary

• The boiling point of a liquid is the temperature at which the vapor pressure of the liquid becomes
equal to the surrounding pressure.

• The normal boiling of a liquid is the temperature at which the vapor pressure of the liquid becomes
equal to 1.00 atmosphere.

Review Questions

1. What happens to the boiling point of a liquid if the pressure exerted on the surface of the liquid is
increased?

2. How can you make water boil without heating it?

3. Fill in the diagram with either "high" or "low" to show how intermolecular forces of attraction influence
the volatility, vapor pressure, and boiling point of a substance.

weak

Wlipti TVTF arp

strona

volatility is

vapor pressure is

boiling point is

vapor pressure is

boiling point is

Vocabulary

boiling point The temperature at which the vapor pressure of a liquid equals the surrounding pressure.

normal boiling point The temperature at which the vapor pressure of a liquid equals 1.00 atmosphere.

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13.5 Heat of Vaporization

Lesson Objectives

• The student will be able to calculate energy changes during phase changes.

• The student will be able to explain the slopes of various parts of heating and cooling curves.

Introduction

In order to vaporize a liquid, heat must be added to raise the kinetic energy (temperature) to the phase
change temperature and then more heat must be added to provide the necessary potential energy to
separate the molecules in the liquid form to the gaseous form.

The Potential Energy Stored in Gases

The difference between the liquid phase and the gas phase of a substance is essentially the distance between
the molecules. Since the molecules attract each other and they are separated by a greater distance in the
gaseous phase than in the liquid phase, the molecules in gaseous phase possess more potential energy
than in the liquid phase. When a substance changes from the liquid phase to the gaseous phase, work
(the physics kind of work as in force times distance) must be done on the molecules to pull them away
from each other. The work done separating the molecules is stored in the molecular structure as potential
energy. If the molecules are allowed to go back together as in the liquid phase, the potential energy is
released - exactly the same amount that was needed to separate the molecules. This potential energy
stored in molecules in the gaseous phase is called the heat of vaporization. The heat of vaporization
(a/Zvap) f° r water is 540 calories/gram (2.26 kJ/g) at the normal boiling point. Because of the strength
of the polar attractions holding water molecules together in the liquid form, water has a fairly high heat
of vaporization. Ammonia, NH3, and ethanol, C2H5OH, which are also polar molecules have heats of
vaporization of 1.38 kJ/g and 0.84 kJ/g respectively.

Example

How much heat in kJ is necessary to vaporize 100. grams of ammonia at its boiling point?

Solution

Heat, Q = (mass)(A// V Ap) = (100. g)(1.38 kJ/g) = 138 kJ

The boiling point of ammonia is — 33°C. It is very important to understand that the ammonia is at the
boiling point before the heat of vaporization is added and after the heat of vaporization is added, the
ammonia is STILL at the boiling point. All the energy involved in the heat of vaporization is absorbed
by the substance as potential energy - none of it goes into kinetic energy and therefore, the temperature
cannot change. A common question asked of students to determine if they understand this point is to ask
which would produce a more severe burn, spilling boiling water at 100°C on your skin or being burned
by gaseous water at 100°C? It may seem at first that since they are both at the same temperature, they
would do the same damage, but in fact, the gaseous water would first deliver a tremendous amount of heat
to your skin as the gas condensed to water (giving up the heat of vaporization) and after you had a burn
from that, then you would have 100°C boiling water on the skin that was already burned.

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Heating and Cooling Curves

The addition of heat before, during, and after a phase transition can be analyzed with the help of a heating
curve. In the heating curve below, a sample of water at 20°C and 1.00 atm pressure had equal quantities
of heat added to it per unit time.

[30-

t /

iao-

ItG-

IOG

VO-

BO-

70-

feO-

-■;

- Jr

30-

ZO-

to-

■ !

Hear Added

Figure 13.12: The heating curve for water at normal pressure.

Between the temperatures of 20°C and 100°C, all the heat added was absorbed as kinetic energy and
therefore the temperature increased. Once the water reached the boiling point, even though the same
amount of heat was added per unit time, the temperature did not increase. Thus, this energy added to the
water DID NOT become kinetic energy. All the heat added to the sample during the time the slope of the
line is zero went into potential energy. This energy represents the heat of vaporization and was used to
do the work of separating the liquid molecules into the gaseous form (greater distance between molecules).
During this flat line period, an observer would see that the water was changing into gas but all of it, both
the part in liquid form and the part in gaseous form, would be at exactly 100°C The temperature cannot
increase until all the liquid has been converted to gas. At 1.00 atm pressure, it is impossible to get water
hotter than 100°C No matter how much heat you add to it, all that happens is that the water boils faster
and converts to gas faster but its temperature will never exceed 100°C It is sort of like the water decides
where the heat goes, and the water decides that ALL the heat added will be used to change phase and
none of it will be used to raise temperature. Once all the water is in the gaseous form, the heat once
again becomes kinetic energy and the temperature of the gas rises. When the gas is cooled, it follows this
same curve in reverse. As it is cooled, the same flat line appears while the heat of vaporization (heat of
condensation now) is removed, and then the temperature may go down again.

Specific Heat

Thermodynamic data (melting point, boiling point, heat of melting, heat of vaporization) for almost
all elements and thousands of compounds are determined by laboratory activity and placed in reference
books. Much of this thermodynamic data is also available now on the internet. Another piece of useful
thermodynamic data about substances is called specific heat. The specific heat for a substance is the
amount of heat required to raise 1.00 gram of the substance by 1.00°C. The symbol C is used for specific
heat and the value for liquid water is 4.18 J/g°C.

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Example

How much heat is required to raise the temperature of 25 grams of water from 15°C to 55°C?

Solution

Q = (mass) (C) (At) = (25 g)(4.18 J/g.°C)(40.°C) = 4180 J = 4.18 kJ

Example

How much heat is required to raise 25 g of water from 25°C to gaseous water at 100°C?

Solution

In this problem, you have to calculate the heat to raise the temperature from 25°C to 100°C and then
calculate the heat of vaporization for the 25 g of water.

Heat raising temp = (mass) (specific heat)(A?) = (25 g)(4.18 J/g.°C)(75°C) = 7840 J = 7.48 kJ
Heat va p 0r i z i n g = (mass)(heat of vaporization) = (25 g)(2.26 kJ/g) = 57 kJ
Total heat = 7.48 kJ + 57 kJ = 64 kJ

Lesson Summary

• The heat of vaporization of a liquid is the quantity of heat required to vaporize a unit mass of that
liquid at constant temperature.

• The energy released when a gas condenses to a liquid is called the heat of condensation.

• The specific heat of a substance is the amount of heat required to raise the temperature of one gram
of the substance by 1.0°C.

Review Questions

1. How much heat is required to vaporize 200. grams of water at 100. °C and 1.00 atm pressure? a//vap
for water is 2.25 kJ/g.

2. How much heat is required to raise 80.0 grams of water from 0°C to 100. °C with no phase change
occurring? The specific heat of water is 4.18 J/g.°C

Further Reading / Supplemental Links

• Chemistry, Matter and Its Changes, Fourth Edition, Chapter 12: Intermolecular Attractions and the
Properties of Liquids and Solids, James E. Brady and Fred Senese, John Wiley & Sons, Inc. 2004.
http : //www . sparknotes . com/testprep/books/sat2/chemistry/chapter5sect ion5 . rhtml

• http : //www . kentchemistry . com/links/matter/heatingcurve . htm

Vocabulary

heat of condensation The quantity of heat released when a unit mass of a vapor condenses to liquid
at constant temperature.

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heat of vaporization The quantity of heat required to vaporize a unit mass of liquid at constant tem-
perature.

Image Sources

(2
(3
(4
(5

(7
(8

(9
(10

(11
(12

Richard Parsons. . CC-BY-SA.

Richard Parsons. The centers of positive and negative charge.. CC-BY-SA.

Richard Parsons. Gas Collection Over Water.. CC-BY-SA.

Richard Parsons. . CC-BY-SA.

Richard Parsons. Standard Distribution Curve. CC-BY-SA.

Richard Parsons. The Three Phases of Matter.. CC-BY-SA.

Richard Parsons. The heating curve for water at normal pressure.. CC-BY-SA.

Richard Parsons. . CC-BY-SA.

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Chapter 14

The Solid State

14.1 The Molecular Arrangement in Solids Con-
trols Solid Characteristics

Lesson Objectives

• The student will describe the molecular arrangement of solids.

• The student will use the molecular arrangement in solids to explain the incompressibility of solids.

• The student will use the molecular arrangement in solids to explain the low rate of diffusion in solids.

• The student will use the molecular arrangement in solids to explain the ability of solids to maintain
their shape and volume.

Introduction

There are many ways to classify solids but the broadest categories are crystalline solids, those with a
highly regular arrangement of their particles, and amorphous solids, those with considerable disorder
in their structures. The regular arrangement of the particles in a crystalline solid produces the beautiful,
characteristic shapes of crystals. These structures are represented by a crystal lattice, a three-dimensional
system of points designating the positions of the component particles (atoms, ions, or molecules). There
are also many important amorphous solids. An example is glass, which is best represented as a solid where
the components were frozen in place before they could get into the organized structure of a crystal. Because
of the disorder in glass, some chemists have referred to glass as a super-cooled liquid.

The Molecular Arrangement in Solids

The molecular arrangement in a solid is one where the molecules (or atoms, or ions) are held in a tightly
packed, highly organized pattern, as in the below figure.

The molecular arrangement in a solid. (Source: Richard Parsons. CC-BY-SA)
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II I

In a solid, the intermolecular forces of attraction have completely overcome molecular motion and the
movement of the particles has been reduced to vibration in place. There are only tiny spaces between the
molecules, not nearly enough space for the particles to move past one another.

The Characteristics of Solids

The intermolecular forces of attraction in solids hold the particles so tightly in place that they cannot
pull away from each other to expand their volume nor can they flow past one another to change shape.
Therefore, solids hold their own shape and volume regardless of their container. There is very little empty
space in the solid structure so solids are virtually incompressible and since molecules can't pass each other
in the structure, diffusion is essentially non-existent beyond the surface layer.

Lesson Summary

• Molecules in a solid maintain both their own shape and their own volume.

• Solids are virtually incompressible and have little diffusion beyond the surface layer.

• The molecular arrangement in solids is a highly organized, tightly-packed pattern with small spaces
and molecular motion reduced to vibration in place.

Review Questions

. Fill

in the types of phase

changes left blank in

the chart below.

Solid -

» Liquid

Liquid

— > Gas

Solid -

» Gas

Sublimation

Liquid

-> Solid

Gas — »

Liquid

Gas — »

Solid

Deposition

14.2 Melting

Lesson Objectives

The student will explain why it is necessary for a solid to absorb heat during melting even though
no temperature change is occurring.

Given appropriate thermodynamic data, the student will calculate the heat required to raise temper-
atures of a given substance with no phase change.

Given appropriate thermodynamic data, the student will calculate the heat required to melt specific
samples of solids with no temperature change.

Given appropriate thermodynamic data, the student will calculate the heat required to produce both
a phase change and temperature change for a given sample of solid.

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Introduction

The melting point of a substance, like its boiling point, is directly related to the strength of the forces
of attraction between molecules. Low melting points are typical of substances whose forces of attraction
are very weak, such as hydrogen gas whose melting point is — 259°C. High melting points are associated
with substances whose forces of attraction are very strong, such as elemental carbon whose melting point
is greater than 3500° C.

Melting Points

Solids, like liquids, have vapor pressure. The vapor pressure of a solid is generally very low because the
forces of attraction in solids dominate molecular motion. The vapor pressure of solids, also like liquids,
increases with temperature. The vapor pressure of liquid water is 760 mm of Hg at 100°C and decreases
as the temperature decreases (non-linearly) to 4.6 mm of Hg at 0°C Solid water (ice) has very low vapor
pressures because of both strong forces holding the molecules together and the fact that our interaction
with ice is usually at low temperatures. For example, at -83°C, the vapor pressure of ice is 0.00025 mm
of Hg. As ice is heated, its vapor pressure increases. At 0°C, the vapor pressure of ice is 4.6 mm of Hg
which also happens to be the vapor pressure of water at 0°C. In fact, for all substances, their solids and
liquids have the same vapor pressure at the melting point. The melting point of a solid is defined as the
temperature at which the vapor pressure of the solid and liquid are the same.

Heat of Fusion

The phase change from solid to liquid, melting, has many similarities to vaporization. The solid must reach
its melting point before the molecules can enter the liquid phase. The molecules in liquid phase, however,
are farther apart than the molecules in the solid phase. Since the molecules attract each other, increasing
the distance between them (from solid structure to liquid structure) requires work. Force must be exerted
to pull the molecules away from each other. The work done in separating the molecules is stored in the
molecules as potential energy in the liquid phase. This is the same process that occurs when the heat of
vaporization must be added to liquid molecules to get them into the gaseous phase. In the case of melting,
this potential energy is called the heat of fusion or the heat of melting.

The word 'fusion' is used several times in science with different meanings. You need to note the context
of the use of the word to know which meaning is intended. In this case, fusion is the name of the liquid to
solid phase change. When a solid melts, the heat of fusion must be added and when a liquid freezes (fuses)
back to solid, the heat of fusion is given off. The heat of fusion for water is 334 Joules/gram.

Example 1

How much heat must be added to 25 grams of ice at 0°C to convert it to liquid water at 0°C?
Solution

Q = (mass)(A// FUS ioN) = (25 g)(334 J/g) = 8350 J = 8.4 kJ

Heating Curves

Phase changes are often analyzed with the help of a heating curve. The heating curve for water at 1.00 atm
pressure appears in Figure 14.1. Many substances will have heating curves similar to the one for water.
The differences will be in the length of flat lines and the slopes of the inclined lines.

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The Heating Curve for Water at 1 .00 atm

120-

i-ij-
[03 "

90-

80-

o^ 70-

| 60-

1 -

30-

20-

10-

2 /

.»-

Figure 14.1: The heating curve for water.

The section of the curve labeled "1" represents the time the sample is completely in the form of ice and is
being heated. All the heat added is going into kinetic energy and so the temperature of the ice is increasing.
The slope of the line is related to the specific heat of ice - the amount of heat required to raise 1.00 gram
of ice by 1.00°C. When the sample reaches 0°C, that is, the melting point, all the heat added starts going
into potential energy. During the section of the graph labeled "2," the sample is being converted from
solid to liquid. Both solid and liquid are present at this time, but all of the sample, both solid and liquid
are at 0°C. No heat will go into kinetic energy until the entire sample is converted to liquid. Therefore,
the temperature remains constant during this period. The total amount of heat added during section "2"
would represent the heat of melting for this sample. Once the entire sample has been converted to liquid,
the added heat would once again go into kinetic energy and the temperature would increase. The slope of
the line in section "3" is related to the specific heat of liquid water. The temperature continues to rise as
heat is added until the boiling point is reached. The temperature again remains constant through section
"4" and the added heat becomes potential energy providing the heat of vaporization. During section "4,"
both liquid water and gaseous water are present. Once the entire sample has been converted to gas, the
added heat again becomes kinetic energy and the temperature of the gas increases. Gaseous water also
has its own specific heat.

Example 2

A 2, 000. gram mass of water in a calorimeter has its temperature raised by 3.00°C while an exothermic
chemical reaction is occurring. How much heat, in joules, is transferred to the water by the heat of reaction?

Solution

The heat is calculated by determining the heat absorbed by the water. This amount of heat is the product
of three factors, 1) the mass of the water, 2) the specific heat of water, and 3) the change in temperature
of the water.

heat = (mass of water) (C water )(At) heat = (2000.g)(4.18 J/g -° C)(3.00°C) = 25,080 joules = 25, 100 joules

Example 3

A 1,000. gram mass of water whose temperature was 50. °C lost 33,440 joules of heat over a 5-minute
period. What was the temperature of the water after the heat loss?

Solution

heat = (mass) (C w ) (At)

heat -33,400 ioules

Af = 7T- = -, b — ; r = -8.00°C

mass x C w (1000. g)(4.18 J/g -° C)

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If the original temperature was 50. °C and the temperature decreased by 8°C, then the final temperature
would be 42°C.

Example 4

Use the thermodynamic data given to calculate the total amount of energy necessary to raise 25 grams of
ice at -20°C to gaseous water at 120°C.

Melting point of ice = 0°C
a//vap for water = 2260 J/g

Boiling point of water = 100°C
A//pusiON fo r water = 334 J/g

C ice = 2.11 J/g.°C

C water = 4.18 J/g.°C

C„ater vapor = 1.84 J/g.°C

Solution

There will be five steps in the solution process.

1. We must calculate the heat required to raise the temperature of the ice from -20°C to 0°C

2. We must calculate the heat required to provide the heat of melting - to change ice at 0°C to water
at 0°C.

3. We must calculate the heat required to raise the temperature of the liquid water from 0°C to 100°C

4. We must calculate the heat required to provide the heat of vaporization - to change liquid water at
100°C to gaseous water at 100°C.

5. Finally, we must calculate the heat required to raise the temperature of the gaseous water from 100°C
to 120°C.

Raising the temperature of ice, Q = mCAt = (25 g)(2.11 J/g.°C)(20°C) = 1051 J

Melting ice to liquid, Q = (mass)(A// FUSI o N ) = (25 g)(334 J/g) = 8350 J

Raising the temperature of liquid, Q = mCAf = (25 g)(4.18 J/g.°C)(100°C) = 10450 J

Vaporizing liquid to gas, Q = (mass)(A//vAp) = (25 g)(2260 J/g) = 56500 J

Raising the temperature of gas, Q = mCAt = (25 g)(1.84 J/g.°C)(20°C) = 920 J

The sum of all five steps is 77, 000 J = 77 kJ

The cooling process would be exact reverse of the heating curve. If water in the gaseous phase is cooled,
each 1.84 Joules of heat removed would lower the temperature of 1.00 g of gas by 1.00°C. When the gaseous
water reaches the boiling point (also the condensation point), each gram of gaseous water that condenses
to liquid will release 2260 Joules of heat. Once all the water is in the liquid form, the removal of each
4.18 Joules of heat by cooling will cause the temperature of 1.00 g of water to cool by 1.00°C At the
freezing point (also the melting point) 334 Joules of heat must be removed to convert each gram of liquid
water to ice. When the entire sample of water is in the form of ice, 2.26 Joules of heat must be removed
to cool each gram by 1.00°C. For all phase changes, going up in temperature or down in temperature, the
entire sample will change phase before any temperature change will occur.

Since both melting points and heats of fusion are dependent on the strength of the attractive forces between
molecules, a solid with a low melting point will usually also have a low heat of fusion and a solid with a
high melting point will have a high heat of fusion.

Lesson Summary

• Solids melt when the vapor pressure of the solid equals the vapor pressure of the liquid.
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• Heat must be absorbed by a solid to become a liquid even though the temperature remains the same.
The quantity of heat absorbed per unit mass is called the heat of fusion.

• Stronger forces of attraction between particles in solids produce higher melting points and higher
heats of fusion.

• During a phase change, all the added energy goes into the heat of fusion (potential energy) and none
goes to kinetic energy (raising temperature).

Review Questions

Use the thermodynamic data given in the Table 14.1 to answer problems 1-5.

Table 14.1: Thermodynamic Data of Various Substances

Water

Cesium, Cs

Silver, Ag

Melting Point
Boiling Point

Affusion

o°c

100.°C
334 J/g

^//vaporization 2260 J/g

Specific Heat, C, for 2.01 J/g ° C

Gas

Specific Heat, C, for 4.18 J/g ° C

Liquid

Specific Heat, C, for 2.09 J/g ° C

Solid

29°C
690.°C
16.3 J/g
669 J/g
0.167 J/g °C

0.209 J/g -° C

0.251 J/g-°C

962°C

2162°C
105 J/g
2362 J/g
0.159 J/g °C

0.294 J/g -° C

0.235 J/g -° C

1. How many Joules are required to melt 100. grams of silver at its normal melting point with no
temperature change?

2. How many Joules are required to boil 150. grams of cesium at its normal boiling point with no
temperature change?

3. How many Joules are required to heat 200. g of liquid water from 25°C to steam at 125°C under
normal pressure?

4. How many Joules are required raise the temperature of 1.00 gram of water from -269°C (the current
temperature of space) to 1.60 x 10 15 °C (the estimated temperature of space immediately after the
big bang)?

5. How many Joules are required to raise the temperature of 1000. g of cesium from -200. °C to +200. °C?

6. Why does the boiling point of water increase with increasing surrounding pressure?

7. Why must heat be absorbed to melt a solid even though both the solid and the liquid are at the
same temperature?

Heating Curve for Water at 1.00 atm Pressure

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8. What is happening to the water in section B?

9. What is happening to the water in section A?

10. Why are the slopes of the lines in sections A, C, and E different?

Vocabulary

crystal A solid consisting of plane faces and having definite shape with the atoms arranged in a repeating
pattern.

freezing The phase change from liquid to solid.

freezing point The temperature at which a liquid changes to a solid.

fusion

1. The change of a liquid to a solid.

2. A nuclear reaction in which two or more smaller nuclei combine to form a single nucleus.

heat of condensation The quantity of heat released when a unit mass of vapor condenses to a liquid
at constant temperature.

heat of fusion The quantity of heat released when a unit mass of liquid freezes to a solid at a constant
temperature.

heat of vaporization The quantity of heat absorbed when a unit mass of liquid vaporizes to a gas at
constant temperature.

joule A basic unit of energy in the SI system, equal to one Newton-meter.

melting The phase change from solid to liquid.

melting point The temperature at which a substance changes from the solid phase to the liquid phase.

14.3 Types of Forces of Attraction for Solids.

Lesson Objectives

• The student describe the metallic bond and explain some of the solid characteristics that are due to
metallic bonding.

• Given the characteristics of a solid such as conductivity of solid and liquid phase, solubility in water,
malleability, and so on, the student will be able to identify the type of solid, i.e. the attractive forces
holding the solid in the solid form.

Introduction

The range of melting points for the various types of solids is extremely wide. Substances with very weak
London dispersion forces like helium, will melt at only a couple of degrees above absolute zero whereas
solid substances like asbestos and diamond do not melt until the temperature is in excess of 3500°C

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Solids Held Together by Intermolecular Forces of Attraction

Each of the intermolecular forces of attraction discussed in the chapter on liquids (London dispersion
forces, polar attractions, and hydrogen bonds) will also produce solids if the temperatures are low enough.
Those intermolecular forces of attraction have the same causes in solids as they do in liquids but because
of the lower temperatures and the closeness of the molecules in solids, the forces will be more effective in
pulling the molecules together. The solids formed due to these intermolecular forces of attraction will be
crystalline solids.

Solids held together by London dispersion forces are not soluble in water, nor are they good conductors
of electricity or malleable. In terms of solubility, the general rule is "like dissolves like." This means that
polar solvents dissolve polar or ionic substances but not non-polar and non-polar solvents dissolve non-
polar solutes but not polar. Since water is a strongly polar molecule, it will dissolve most polar solids.
Polar solids are not good conductors and are not malleable.

Ionic Solids

Ionics solids are held together by the electrostatic attraction between oppositely charged ions. The ions
are formed into various types of crystal lattice structures depending on the comparative sizes of the ions
and the charges on each. These ionic charges are full 1+, 2+, 1-, 2-, and so on, charges and so they
are considerably stronger than either polar attractions or the especially strong hydrogen bonds. This will
cause the melting points for ionic substances to be quite high compared to the substances we have been
considering. For example, the melting point of sodium chloride, NaCl, is 801°C and the melting point of
calcium sulfate, CaSO^, is 1460° C.

In solid state, the ions in ionic solids are held firmly in position and there are no spaces large enough for
the ions to move through even if they could escape the forces of attraction. Since the ions cannot move,
ionic solids are non-conductors of electricity. When the solid is melted to a liquid, however, the ions are
free to migrate and therefore, ionic liquids are good conductors of electric current.

In ionic crystals, the positive and negative ions occupy positions so the ions will attract the maximum
possible number of oppositely charged ions. (See Figure 14.2).

Figure 14.2: A crystal.

In the NaCl lattice, each sodium ion is equally attracted to six chloride ions. In the sketch you can see
each sodium surrounded by four chlorides, and in a three-dimensional structure, there would be another
chloride above the sodium ions in the layer above and another one in the layer below, for a total of six.

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The word malleable (in science) refers to a substance that can be pounded or beaten into another shape
without breaking. Ionic solids are NOT malleable. (See Figure 14.3).

Figure 14.3: Ionic solids are not malleable.

If an ionic crystal is struck, the ions are driven down a layer so that they will be next to like charges.
These charges will repel and the crystal will shatter.

Ionic solids are usually quite soluble in water and the water solutions of ionic solids are good conductors
of electricity because of the freedom of the ions to migrate through the solution.

Metallic Solids

Of the 81 elements which can be clearly classified as metals, all of them except mercury are solids at
room temperature. Any model to explain the bonding in metallic solids must account for the properties
of metals, some of which are quite unusual. The properties of metals include 1) excellent conductors of
heat and electricity in both solid and liquid phase, 2) malleable, 3) most of them are white and shiny, 4)
metallic solids are not soluble in any common solvent, polar or non-polar, and 5) they have a wide-range
of melting points mostly higher than the melting points of polar solids.

The simplest bonding model that has been proposed to explain metallic behavior is the metallic bond,
which envisions a regular pattern of cations surrounded by a "sea of electrons." The metal ions (all of
the metal atom except the valence electrons) occupy positions in a lattice structure while the mobile,
free-moving, sea of valence electrons occupy all of the overlapping valence shell area. See below figure.

The metallic bond consists of non-directed bonds in which a "sea of electrons" surrounds all the bonded
atoms. (Source: Richard Parsons. CC-BY-SA)

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All the atoms are bonded in a single bond that includes the entire piece of metal. The bond is non-
directional and does not hold adjacent ions in position relative to each other.

For purposes of comparison, consider the covalent bonding in a trigonal planar molecule below. The central
atom in this molecule contains three pairs of electrons. The electrons take positions as far away from each
other as possible due to electrostatic repulsion. Therefore, the pairs of electrons maintain directed positions
at angles of 120° from each other. The atoms that share these electron pairs in the covalent bond must be
placed so that the shared electrons are in the overlapped orbitals of both atoms. Therefore, these bonded
atoms may not move with respect to each other. The atoms hold their relative positions because of the
directional bonds. Neither the bonding electrons nor the atoms are free to move with respect to each
others.

Directional bonding in trigonal planar molecule. (Source: CK-12 Foundation. CC-BY-SA)

The model of the metallic bond, however, provides for mobile electrons that are free to move throughout
the entire piece of metal and therefore, provides the means for the metal to be an excellent conductor of
heat and electricity. Extra electrons pushed onto the metal on one side can easily move through the valence
electron shells to the other side. The metal ions are not directionally bonded to their immediate neighbors
and this allows them to be pushed to new positions without breaking the bond. As long as an atom or ion
is not separated from the piece of metal, its position can be significantly changed while it remains bonded
to the other atoms. This malleability allows metal cubes to be pounded into flat sheets without breaking
the bond.

The freedom of the electrons on the surface of a piece of metal also allows the metal to absorb and emit
many frequencies of light which accounts for the white, shiny appearance of many metals. The metals
on the far left of the periodic table have the fewest valence electrons in the valence shells so the valence
electrons in these metals would be least crowded, have the most freedom, and present the most complete
metal character. The metals of families IA and IIA are excellent conductors, exceptionally malleable (soft
enough to be cut with a spoon), and white and shiny in color.

Other elements can be introduced into a metallic crystal relatively easily to produce substances known as
alloys. An alloy is defined as a substance that contains a mixture of elements and has metallic properties.
A fairly well-known alloy is brass which is an alloy composed of about two-thirds copper atoms and one-
third zinc atoms. Sterling silver is an alloy composed of about 93% silver and 7% copper.

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Iron is a metal that is commonly alloyed with carbon to produce steel. The carbon forms directional
bonds with some of the iron atoms to make steel less malleable than pure iron. Steel with less than 0.2%
carbon remains somewhat malleable and is used for nails and cables. Steel with around 0.6% carbon is
harder and is used for railroad rails and structural steel beams. Steel with around 1.5% carbon is very
hard and is used for tools and cutlery.

Malleability, ductility, and conductivity are examined, along with methods for extracting metals from ores
and blending alloys. Metals (http: //www. learner. org/vod/vod_window.html?pid=811)

Network Solids

In some solids, all the atoms in the entire structure are bonded with covalent chemical bonds. These
solids are a single giant molecule and are called network solids. When considering the strength of the
bonds/attractions that hold various particles together in the solid state, the strongest of them all are
covalent bonds. Therefore, network solids have the highest melting points of all solids. To melt a network
solid requires enough molecular motion to disrupt covalent chemical bonds. Network solids are not soluble
in any common solvent. Most network solids are non-conductors but graphite is an exception, it is a good
conductor of electricity. Some examples of network solids are graphite, diamonds, mica, and asbestos.

The solid structure in graphite involves large two-dimensional molecules of covalently bonded carbon atoms.
The carbon atoms form flat sheets (like a sheet of paper) bonded in the fashion shown in Figure 14.4.

Graphite: a two-dimensional network solid.

/N/N/N

c c c

e \,/ c \ e / e \ e /%

Diamond: a three-dimensional network solid.

\ c / C \c/ e \,/ <

"<5

\</ c \ e /'

Figure 14.4: One layer (sheet) of graphite, a dimensional network solid, and diamond, a dimensional
network solid.

Then layers of these sheets are laid on top of each other and the sheets are held together by much weaker
London dispersion forces. The sheets are extremely strong in the two dimensions involving covalent bonds
but the forces holding the sheets together are weak and easily broken. The flat sheets slide over each other
readily and this makes graphite a good lubricant for metal parts.

The mineral mica is also bonded in this two-dimensional network style. Mica is found in nature and appears
as a rock but you can slide your fingernail between sheets and pull off large flats sheets of the rock. One
type of mica, called muscovite mica, is transparent enough that you can see through several sheets. This
material has been used to make small windows in furnaces so the operator can look in but the window is
rock, not glass, so it won't melt.

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Diamonds are giant molecules of carbon atoms bonded three-dimensionally in tetrahedral units. Every
carbon atom in the structure is covalently bonded to four other carbon atoms. Diamonds are the hardest
substance known and have one of the highest melting points of all substances.

Some forms of asbestos are a one-dimensional network solid in which atoms are bonded in a chain. The
result is a fibrous molecule that can be woven into fabric. (See Figure 14.5).

Figure 14.5: Asbestos in the form of chrysotile. You can see fibers of asbestos in the upper left hand corner
of the photo.

Due to the high melting point, asbestos fabric was used to make heat resistant materials (fireman's gloves,
furnace padding, clutch plates, etc.) for many years until it was determined that asbestos fibers are
hazardous if inhaled.

The following web site has data and explanatory reasons for the trends in melting and boiling points
of some period 3 elements. Trends in Melting Points and Boiling Points for Period 3 elements (http:
//www. creative-chemistry.org.uk/alevel/modulel/trends8.htm)

Amorphous Solids

Many important solids do not have the regular, repeating arrangement of atoms or molecules that is present
in crystalline solids. Solids with irregular, unpredictable molecular organization are called amorphous
solids. There are many solids that will form either crystalline or amorphous solids depending on how
rapidly the liquid is cooled. Very rapid cooling of these substances frequently results in an amorphous
solid whereas slow cooling produces crystalline solids. Amorphous solids have been described as appearing
to have their molecules frozen in place before they had time to get into an organized pattern. Examples
of amorphous solids are glass, paper, plastics, cement, and rubber.

Amorphous solids are called solids because they maintain their shape and volume. Some researchers insist
that certain amorphous solids will flow under pressure, which is a characteristic of liquids. Antique windows
have been found that are thicker at the bottom than at the top. Some chemists claim this is because the
glass very slowly, over a hundred years, flowed downward under gravitational force. Other chemists claim
the antique glass being of different thicknesses was caused by flaws in the glass making process of a hundred
years ago. There was no mention in the opinion of why the thicker part of the glass was always at the
bottom and never at the top.

Crystalline solids melt at sharply defined temperatures. Glass and some other amorphous solids, however,
soften as they are heated. Some authors refer to amorphous solids as "not true solids", others call them
"supercooled liquids," and still others insist that amorphous solids are absolutely solids. The importance
of amorphous solids such as glass and plastics will insure that research on their structure continues.

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Lesson Summary

• One type of solid is formed by ionic solids in which the inter-particle forces of attraction are electro-
static attractions due to the opposite charges of the ions.

• One type of solid is formed by metallic atoms where a sea of electrons exerts a force of attraction on
the positive ions (metallic bond).

• Network solids have every atom in the structure attached to other atoms in the structure by covalent
chemical bonds.

• Amorphous solids are solids that cooled so rapidly, the molecules did not get into the tight, organized
solid pattern. Due to their disorganized structure, amorphous solids have some properties more like
liquids.

Review Questions

1. Identify the most important type of inter-particle force present in the following solids that is respon-
sible for binding the particles into a solid.

(a) He

(b) NO

(d) BaCh

(e) CH A

(f) NaN0 3

(g) co 2

(h) CHCh

(i) pure Mg

(j) diamond

2. Predict which substance in the following pairs would have the stronger force of attraction between
molecules and justify your answer.

(a) C0 2 or OCS

(b) PF 3 or PF 5

(d) H 2 or H 2 S

(e) solid argon or solid sodium

(f) HF or HBr

3. In the following groups of substances, pick the one that has the requested property and justify your
answer.

(a) highest boiling point: HCl,Ar,F 2

(b) highest melting point: H 2 0,NaCl,HF

4. An unknown solid is not soluble in water or CC/4. The solid conducts electricity and has a melting
point of 800°C Identify the most likely attractive forces holding the particles in the solid state.

5. An unknown solid is soluble in water but not in CC/4. The solid does not conduct electricity but
its liquid does. The solid shatters when hammered and has a melting point of 1430°C. Identify the
most likely attractive forces holding the particles in the solid state.

6. Why would you expect ionic solids to have higher melting points that polar solids?

7. Why does the melting point of water decrease with increasing surrounding pressure?

8. List the following substances in order of increasing boiling points: BaCl 2 , H 2 , CO, HF, Ne, C0 2 .

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(a) H 2

(b) CH-iOH

(d) KCl

(e) CO

Select your answers for questions 9, 10, and 11 from these choices.

9. Which of these substance is most likely to be a solid at 25°C and 1.0 atm?

10. Which of these substances is capable of hydrogen bonding?

11. Which of the substances has its solid properties governed by London dispersion forces?

12. Place these molecules, CF4, CaCh, and ICl, in order of decreasing melting points (highest first).

(a) CFi > CaCh > ICl

(b) CaCh > ICl > CF A

(d) ICl > CF 4 > CaCh

(e) CF A > ICl > CaCh

Further Reading / Supplementary Links

• http : //learner . org/resources/series61 . html

Website with lessons, worksheets, and quizzes on various high school chemistry topics.

• Lesson 4-6 is on Intermolecular Forces.

• Lesson 210 is on Heat Transfer Calculations.

• http : //www . f ordhamprep . org/gcurran/sho/sho/lessons/lesson46 . htm

• http : //dwb . unl . edu/teacher/nsf/cO 11 inks/www . ualberta . ca/ ! bderksen/f lor in . html

• http : //www . sciencedaily . com/releases/2008/07/080704153507 . htm

• http : //www . bestcrystals . com/crystals2 . html

Vocabulary

alloy A substance composed of a mixture of two or more elements and having metallic properties.
conductivity The property of being able to transmit heat and/or electricity.
conductor A substance that can transmit heat and/or electricity.
ductility The property of a substance that allows it to be drawn into a wire.
electrical conductivity The ability of a substance to transmit an electric current.
malleable The property of being able to be hammered or rolled into sheets.

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metallic bond The attractive force that binds metal atoms together. It is due to the attractive force
that the mobile electrons exert on the positive ions.

specific heat The amount of energy necessary to raise 1.00 gram of a substance by 1.00°C.

14.4 Phase Diagrams

Lesson Objectives

• The student will be able to read specific requested information from a phase diagram.

• The student will be able to state the primary difference between a generic phase diagram and a phase
diagram for water.

Introduction

A phase diagram is a convenient way of representing the phase of a substance as a function of temperature
and pressure. Phase diagrams are produced by altering the temperature of a pure substance at constant
pressure in a closed system. This process is repeated at many different pressures and the resultant phases
charted.

Generic Phase Diagram

The phase diagram in Figure 14.6 is a generic phase diagram that would be produced by many pure
substances. The differences in the diagram for different substances would be in the specific thermodynamic
points like melting points, boiling points, and so on and differences in the slopes of the curved lines. The
general shape of the phase diagram would be very similar for many substances.

g x

Solid

/ Liquid

T? r

Gas

\^ s

Temp«petuj*« in °C

Figure 14.6: A generic phase diagram.

In the pink area in the diagram, the substance would be in the solid state, the purple area represents liquid,
and in the yellow area, the substance would be gaseous. Following a constant pressure line, such as XY,
shows the phase of the substance at different temperatures for this pressure. Since line XY crosses from
solid into liquid at point A, this temperature (B) would be the melting point of the substance. Continuing
along the line, we see it crosses from liquid to gas and that point corresponds with temperature C. This is
the boiling point of the substance at pressure X. The line between the pink and purple areas represent the
various melting points at different pressures and the line separating the purple area from the yellow area

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represents the boiling point at various pressures. At the melting points, both solid and liquid can exist at
the same time as the phase changes occurs but on either side of this temperature, the substance must be
all in one phase. At the boiling points, the substance may exist in both liquid and gas phase at the same
time but only exactly at the boiling point. There is one point on the diagram where all three phases may
exist at one time. This point is called the triple point. The pressure at this point is called the triple
point pressure and the temperature at this point is called the triple point temperature.

There is also a line separating the pink area from the yellow area. This line represents the phase change
in which a solid changes directly to a gas without passing through the liquid phase. This phase change is
known as sublimation. All substances undergo sublimation at the appropriate pressures. We do not see
sublimation often because the pressures are frequently quite low and we do not encounter substances at low
pressures in our daily lives. Some of us have seen carbon dioxide, CO2, in the solid form which is called dry
ice. If you have seen dry ice, you have noticed that the substance goes from the solid phase to the gaseous
phase at room conditions without passing through a liquid phase. In the phase diagram for dry ice, we
would see that the triple point is above normal atmospheric pressure and so at room conditions, carbon
dioxide undergoes sublimation. Water also undergoes sublimation but the pressure where this would occur
is below 4.58 torr or 0.0060 atm - a pressure we seldom witness.

Solid

Liquid

A i

TP/

Gas

Temperature in C

Figure 14.7: Another generic phase diagram.

Figure 14.7 shows the same generic phase diagram we looked at before. Two points have been added to
the diagram, labeled A and B. You should note that the substance at point A can be caused to go through
a phase change from solid to gas (sublimation) in two different ways. The substance could be heated at
constant pressure or the substance could undergo a lowering of pressure at constant temperature. Both of
these procedures would cause the solid to undergo sublimation. Point B is a similar circumstance except
that the substance begins as a liquid. The liquid at point B could be caused to under a phase change to
gas either by heating it at constant pressure or by lowering the pressure at constant temperature. You
might also note that the substance at the triple point will become a solid if pressure is increased and will
become a gas if pressure is decreased.

Phase Diagram for Water

The phase diagram for water has one very interesting difference from the generic phase diagram. Please
note that this diagram is not drawn to scale. If the distance between 1.0 atm and 218 atm was drawn to
scale, the difference between 1.0 atm and 0.006 atm would be invisible. The diagram is drawn just to show
specific points of interest. (See Figure 14.8).

The primary difference in the shape of this diagram and the generic diagram is that the solid-liquid
equilibrium line (melting point) has a negative slope (tilts backwards). The positive slope of this line in
the generic diagram indicates that as pressure increases, the melting point increases. That is reasonable

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Phase Diagram for Water
213

£1-

1.0

0.006

374

Temperature in C

Figure 14.8: The phase diagram for water.

because more pressure on the surface would require a higher temperature to overcome that extra pressure
and melt the substance. The negative slope of this line in the water diagram indicates that as the pressure
increases, the melting point of water decreases. The reason this occurs is because the increased pressure
breaks some of the hydrogen bonds in the water and so LESS temperature is needed to melt ice at higher
pressures. There is plenty of evidence of this in everyday life if you look for it. We all think of ice as being
a very slippery substance but the surface of ice is no different from the surface of many other solids. The
surface of ice is not smoother than the surface of copper or glass or most other solids. The reason that
we slip on ice is that when you stand on ice or drive your car on ice, the pressure of your weight or the
weight of your car causes the ice to melt. When you put pressure on ice, the surface between the ice and
the weight is liquid and that makes it slippery. If you are an ice skater, you are aware that when you look
closely at the track of the blade of an ice skate on ice, the track is filled with liquid, not solid.

If you follow the line at 1.0 atm pressure for water, you see that the melting points and boiling points are
the temperatures noted as the normal melting point and the normal boiling point. The triple point for
water is at 0.006 atm and 0.0098°C Very expensive equipment is necessary if you wish to see water at
its triple point. There are commercial processes that make use of the sublimation of water. Foods that
are referred to as "freeze dried" have the surrounding pressure and temperature reduced to a point below
the triple point and then they are heated while a vacuum pump removes vapor to keep the pressure below
the triple point pressure. This causes the water in the food to sublimate (solid to gas) and it is drawn off
by the pump. The end result is the food minus all the water that was in it. This process is supposed to
damage the food less than getting rid of the water by heating the food. The idea is that you can just add
water and the food will be like it was originally. The success or failure of that idea is for you to decide.

As the temperature of liquid is raised, the amount of pressure that is required to squeeze the substance and
keep it in liquid form also increases. Liquids will eventually reach a temperature at which no amount of
pressure will keep it in the liquid form. The substance at that temperature will vaporize regardless of the
amount of pressure on it. The highest temperature a liquid reaches and can still be squeezed into a liquid
is called the critical temperature. The pressure that is required at the critical temperature to force the
liquid to stay in liquid form is called the critical pressure. The critical temperature and pressure for
water is 374°C and 218 atm.

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Lesson Summary

Phase diagrams are graph showing the phase of a substance at various conditions of temperature and

pressure.

The phase diagram for water is different from most phase diagrams because unlike most substances,

the melting point of water decreases as pressure increases.

Review Questions

1. Consider the phase diagram below.

Name the phases that may be present at each lettered point in the diagram.

Further Reading / Supplementary Links

• Chemistry, Matter and Its Changes, 4 th Edition, Chapter 12: Intermolecular Attractions and the
Properties of Liquids and Solids, James E. Brady and Fred Senese, John Wiley &; Sons, Inc., 2004.
http : //dwb . unl . edu/teacher/nsf/cO 11 inks/www . ualberta . ca/ ! bderksen/f lorin . html

• http : //www . sciencedaily . com/releases/2008/07/080704153507 . htm

Vocabulary

critical pressure The pressure required to liquefy a gas at its critical temperature.

critical temperature The highest temperature at which it is possible to liquify the substance by in-
creasing pressure.

Image Sources

(1) http://www.openchemistry.co.uk/images/diamond. Creative Commons Attribution 3.0.

(2) Richard Parsons. Ionic solids are not malleable.. CC-BY-SA.

413 www.ckl2.ors

(3) Richard Parsons. The phase diagram for water.. CC-BY-SA.

(4) Richard Parsons. A NaCl crystal. CC-BY-SA.

(5) Richard Parsons. The heating curve for water.. CC-BY-SA.

(6) Richard Parsons. A generic phase diagram.. CC-BY-SA.

(7) http://en.wikipedia.0rg/wiki/File:AsbestoslUSGOV.jpg. Public Domain- USGov.

(8) Richard Parsons. Another generic phase diagram.. CC-BY-SA.

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Chapter 15

The Solution Process

15.1 What Are Solutions?

Lesson Objectives

• Define solutions.

• Describe the composition of homogeneous solutions.

• Describe the different types of solutions that are possible within the three states of matter.

• Identify homogeneous solutions of different types.

Introduction

In this chapter we begin our study of solution chemistry. We have previously discussed a number of the
concepts you will be learning in this chapter. But we will explore these ideas in greater detail in this
chapter. Let's begin with a discussion of the definition of a solution. We all probably think we know what
a solution is. We might be holding a can of soda or a cup of tea while reading this lesson and think ...
hey this is a solution. Well, you are right. But you might not realize that alloys, such as brass, are also
classified as solutions, or that air is a solution. Why are these classified as solutions? Why wouldn't milk
be classified as a true solution? To answer these questions, we have to learn some specific properties of
solutions. Let's begin with the definition of a solution and view some of the different types of solutions.

Homogeneous Mixtures

A homogeneous mixture is a solution of the same appearance or composition throughout. Thinking of
the prefix "homo" meaning "sameness", this definition makes perfectly good sense. Homogeneous solutions
carry the same properties throughout the solution. Take, for example, vinegar that is used in cooking.
Vinegar is approximately 5% acetic acid in water. This means that every teaspoon of vinegar that is
removed from the container, contains 5% acetic acid and 95% water.

A point should be made here that when a solution is said to have uniform properties throughout, the
definition is referring to properties at the particle level. Well, what does this mean? Let's consider brass
as an example. The brass is an alloy made from copper and zinc. To the naked eye this brass coin seems
like it is just one substance but at a particle level two substances are present (copper and zinc). An
alloy is a homogeneous solution formed when one solid is dissolved in another. So the brass represents a

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homogeneous mixture. Now, consider a handful of zinc filings and copper pieces. Is this now a homogeneous
solution? The properties of any scoop of the "mixture" you are holding would not be consistent with any
other scoop you removed from the mixture. Thus the combination of zinc filings and copper pieces in a
pile does not represent a homogeneous solution. Another example of a solution is margarine. Margarine is
a combination of a number of substances at the molecular level but to the naked eye it is a homogeneous
solution that looks like just one substance.

Varying Concentrations of Ingredients Produces Different Solu-
tions

The point should be made that because solutions have the same composition throughout does not mean
you cannot vary the composition. If you were to take one cup of water and dissolve 1/4 teaspoon of table
salt in it, a solution would form. The solution would have the same properties throughout, the particles of
salt would be so small that they would not be seen and the composition of every milliliter of the solution
would be the same. But you can vary the composition of this solution to a point. If you were to add another
1/2 teaspoon of salt to the cup of water, you would make another solution, but this time there would be a
different composition than the last. You still have a solution where the salt particles are so small that they
would not be seen and the solution has the same properties throughout, thus it is homogeneous. What
would happen if you tried to dissolve 1/2 cup of salt in the water. Would the solution stay homogeneous?
No, it would not. At this point, the solution has passed its limit as to the amount of salt it can dissolve
and it would no longer be a homogeneous solution.

So solutions have constant composition but you can vary the composition up to a point. There are limits
to the amount of substance that can be dissolved into another substance and still remain homogeneous.

Types of Solutions

There are three states of matter: solid, liquid, and gas. If we think about solutions and the possibilities of
combining these states together to form solutions, we have nine possibilities. Look at the Table 15.1.

Table 15.1: Types of Solutions

Solid Liquid Gas

Solid Solid in a Solid Solid in a Liquid Solid in a Gas

Liquid Liquid in a Solid Liquid in a Liquid Liquid in a Gas

Gas Gas in a Solid Gas in a Liquid Gas in a Gas

In the table (15.1), there are really only four that are common types of solutions. These are shown in
boldface. The others, although still solutions, are less common in everyday lives. For example, a solid
in a liquid solution can be anything from salt or sugar solution, to seawater. Liquid in liquid solutions
include the antifreeze/coolant we use for our cars and vinegar. For a gas in a liquid solution, the most
common example is soda pop: carbon dioxide dissolved in water (with lots of sugar!) Another example is
the ammonia solution you may use (or have seen used) to clean in the home. Finally, to understand the
gas in a gas solution, take a deep breath. That's right, air is a solution made up of mostly oxygen gas and
nitrogen gas.

A solid in a solid solution is less common but still we see a lot of steel and brass around in our everyday
world. These are examples of solid - solid solutions. The other types of solutions are less common but do
exist in the world of solution chemistry.

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Lesson Summary

• A solution is a mixture that has the same properties throughout. Solutions have the same composition
throughout but this composition can vary up to a point, or limit.

• With three states of matter, four types of solution can be classified as the most common as far as
occurring in the everyday world. These include solid in a liquid, liquid in a liquid, gas in a liquid,
and gas in a gas. The other types are less common.

Review Questions

1. What makes a solution homogeneous?

2. Which of the following are homogeneous? Explain.

(a)

gasoline

(b)

chocolate

(c)

blood

(d)

brass

Which of the following is

a solution?

(a)

milk

(b)

blood

(c)

gold

(d)

air

(e)

sugar

4. Which of the following is not a true solution?

(a) vinegar

(b) sand and water

(d) mercury alloy

5. Give an example of a homogeneous solution that is made from the following combinations

(a) a gas in a liquid

(b) a solid in a solid

(d) a gas in a gas

6. Jack is practicing some household chemistry. He takes 1 tsp of sugar and dissolves it in 250mL of
water. He sees that the solution remains clear so continues his experiment by adding a second tsp of
sugar. Stirring the solution makes this solution turn clear. After a few more attempts, Jack sees the
solution turn murky then sugar crystals sinking to the bottom. What is Jack demonstrating?

Further Reading / Supplemental Links

• http://en.wikipedia.org

Vocabulary

solution A homogenous mixture; composition can vary; but composition is the same throughout once
the solution is made.

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15.2 Why Solutions Occur

Lesson Objectives

• Describe why solutions occur; the "like dissolves like" generalization.

• Determine if solutions will occur by studying the molecular structure.

• State the importance of water as the "universal solvent."

Introduction

We have learned that solutions can be formed in a variety of combinations using solids, liquids and gases.
We also know that solutions have constant composition and we can also vary this composition up to a
point to maintain the homogeneous nature of the solution. But how exactly do solutions form? Why is it
that when you mix oil and water together a solution does not form even though it is a liquid in a liquid
and yet vinegar and water will? Why could we dissolve table salt in water but not in vegetable oil? The
reasons why solutions form will be explored in this lesson, along with a discussion of why water is used
most frequently to dissolve substances of various types.

Here is some new vocabulary you will meet in this chapter.

Solution: a homogeneous mixture of substances

Solute: the substance dissolved in a solution, usually determined by being the smaller quantity

Solvent: the substance the solute is dissolved in, usually determined by being the larger quantity

Similar Structures Allow Solutions to Occur

Over the course of your study in chemistry you have learned the terms polar and non-polar. Recall that
in chemistry, a polar molecule is one that has a positive end and a negative end while nonpolar molecules
have charges that are evenly distributed throughout the molecule. In fact, during the study of Valence
Shell Electron Pair Repulsion Theory (VSEPR), you learned that the chemical structures themselves have
built in molecular polarity.

In solution chemistry, we can predict when solutions will form and others won't using a little saying ... "like
dissolves like". The "like dissolves like' saying helps us to predict solubility based on the two parts of a
solution having similar intermolecular forces. For example, suppose you are dissolving methanol in water.
Both methanol and water are polar molecules and form a solution because they both have permanent
dipoles (positive and negative parts of the molecules) that allow the molecules of each of the substances to
be attracted to the other. When this occurs, a solution is made.

A way to understand this is to think about why Velcro is used to hold two different pieces of fabric together.
The two sides of Velcro allow the pieces of fabric to be fastened together because the Velcro has similar
structure that "attract" each other. However, one side of Velcro would not stay together with a piece of
silk since the silk doesn't have any part of its structure with which the Velcro can connect.

Let's look at the individual structure of the water and methanol molecules. Notice in the representation
of the individual molecules of methanol and water how the methanol has a permanent dipole due to the
C — O — H bonds and is a polar molecule. In the representation of the water molecule, you can see that
there are also permanent dipoles making it a polar molecule. The intermolecular forces for both of these
molecules are dipole-dipole attractions. Since these molecules are both polar, they will form a solution
when mixed together. We say they are miscible, which means these two liquids will make a solution. (See
Figure 15.1).

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4H^

sodium chloride crystal

hydrated sodium ion

hydrated chloride ion

The more electronegative oxygen atom pulls the shared
electrons away from the hydrogen atoms causing an
unequal distribution of electrons over the water
molecule. The hydrogen end of the water molecule will
be slightly negative and the oxygen end will be slightly
positive. A molecule with this permanent uneven
distribution of electrons is said to be polar.

When a polar or ionic compound is introduced into
water : ionic charges on the ions or the poles on a polar
molecule are attracted to the poles on the water
molecule and the substance dissolves.

Figure 15.1: The hydration of ions in a polar solvent.

The same is true for the case of a non-polar substance such as carbon tetrachloride being dissolved in
another nonpolar substance such as pentane. London-dispersion forces are the intermolecular bonds that
hold the carbon tetrachloride together as a liquid. London-dispersion forces are also the forces that allow
pentane to be a liquid at room temperature. Since both of these substances have the same intermolecular
forces, when they are mixed together, a solution will be formed.

Unlike the polar molecules, the non-polar molecules have, at any given time, no permanent dipole. If we
were to add table salt, NaCl, to either carbon tetrachloride or pentane, we would find that the salt would
not dissolve. The reason for this is again explained with the structures of the substances. Since carbon
tetrachloride (or pentane) has no permanent dipoles in its molecules, there would be place for the charges
particles in a crystal of NaCl to be attracted.

In a polar solvent, the molecules of solvent are attracted to each other by the partial charges on the ends
of the molecules. When a polar solute is added, the positive polar ends of the solute molecules attract the
negative polar ends of the solvent molecules and vice versa. This attraction allows the two different types
of molecules to form a solution. If a non-polar solute was added to a polar solvent, the non-polar solute
particles cannot attract the solvent molecules away from each other - so a solution does not form.

Polar solvents will dissolve polar and ionic solutes because of the attraction of the opposite charges on the
solvent and solute particles. Non-polar solvents will only dissolve non-polar solutes because they cannot
attract the dipoles or the ions.

Water: The Universal Solvent

Think of the title of this section, water: the universal solvent. The term solvent is used to represent the
medium that is used to produce the solution. The term universal is used to describe the fact that water,
along with many of its other unique aspects, can dissolve many types and kinds of substances. For instance,
table salt, NaCl, is an ionic compound but easily makes a solution with water. This is true for many ionic
compounds. And from your own experience you know that table sugar, a polar covalent compound, also
dissolves in water. And this is also true for other polar compounds such as vinegar and corn syrup.

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Even some nonpolar substances dissolve is water but only to a limited degree. Have you ever wondered
why fish are able to breathe? Oxygen gas, a nonpolar molecule, does dissolve in water and it is this oxygen
that the fish take in through their gills. Or, one more example of a nonpolar compound that dissolves in
water is the reason we can enjoy carbonated sodas. Pepsi-cola and all the other sodas have carbon dioxide
gas, CO2, a nonpolar compound, dissolved in a sugar-water solution. In this case, to keep as much gas in
solution as possible, the sodas are kept under pressure. But that's another part of the story!

Lesson Summary

• Whether or not solutions are formed depends on the similarity of polarity or the "like dissolves like"
rule. Polar molecules dissolve in polar solvents, non-polar molecules dissolve in non-polar solvents,
and ionic molecules in polar solvents. Polarity is determined from molecular geometry.

• Water is considered as the universal solvent since it can dissolve both ionic and polar solutes, as well
as some non-polar solutes (in very limited amounts).

Combinations to form solutions with polar, ionic, and non-polar substances in Table 15.2.

Table 15.2:

Combination Solution formed

Polar substance in a polar substance yes

Non-polar substance in a non-polar substance yes

Polar substance in a non-polar substance no

Ionic substance in a polar substance (i.e. water) yes

Ionic substance in a non-polar substance no

Review Questions

1. What is the "like dissolves like" generalization and provide an example to illustrate your answer.

2. Why will LiCl not dissolve in CC/4?

3. Will acetic acid dissolve in water? Why?

4. What is the difference between intermolecular and intramolecular bonds?

5. In which compound will benzene (CqHq) dissolve?

(a) Carbon tetrachloride

(b) water

(d) none of the above

6. In which compound will sodium chloride dissolve?

(a) Carbon tetrachloride

(b) methanol

(d) none of the above

7. In which compound will ammonium phosphate dissolve?

(a) Carbon tetrachloride

(b) water

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(d) None of the above

8. Thomas is making a salad dressing for supper using balsamic vinegar and oil. He shakes and shakes
the mixture but cannot seem to get the two to dissolve. Explain to Thomas why they will not
dissolve.

Further Reading / Supplemental Links

• http://en.wikipedia.org

Vocabulary

intermolecular bonds Forces of attraction between molecules.
intramolecular bonds Forces of attraction between atoms in a molecule.
universal solvent A solvent able to dissolve practically anything (water).

15.3 Solution Terminology

Lesson Objectives

• Define solute, solvent, soluble, insoluble, miscible, immiscible, saturated, unsaturated, concentrated,
and dilute.

Introduction

Like any discipline that you would study or learn in our world, there are terms that are a part of the
normal day- to- day conversation that takes place. The same is true when people get together to talk
about solutions. You cannot talk about skateboarding without knowing about grinds and slides; you can't
talk about bowling without knowing about spares and strikes; and, you can't talk about solutions without
knowing terms such as solute, dilute, and saturated, just to name a few. In this section, you will learn the
terms of solution chemistry.

Solvent and Solute

The solvent and solute are the two basic parts of a solution. The term solvent is the substance present
in the greatest amount. The solute, then, is the substance present in the least amount. Let's think for a
minute that you are making a cup of hot chocolate. You take a teaspoon of chocolate and dissolve it in
one cup of hot water. Since the chocolate is in the lesser amount it is said to be the solute; and the water
is the solvent since it is in the greater amount.

Sample question: Name the solute and solvent in each of the following mixtures

(a) Kool-aid

(b) iced tea

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Solutions:

(a) solute = Kool-aid crystals (and sugar); solvent = water

(b) solute = ice tea mix (and sugar); solvent = water

Soluble and Insoluble

Did you ever try to make a solution and yet it did not matter how much shaking and stirring that you
did, the solid just kept falling to the bottom of the container. Try it for yourself. Take some sand and try
to dissolve it in a cup of water. What happens? The sand will not dissolve. It is insoluble. Now if you
were to take a teaspoon of table salt or sugar and do the same experiment, what a different result. There's
no problem dissolving these substances. Salt and sugar are both soluble in water. When a substance is
soluble, it means that the substance has the ability to dissolve in another substance. And being insoluble
means that the substance does not dissolve.

Miscible and Immiscible

When referring to liquid solutes in liquid solvents, we can use the terms miscible and immiscible. Liquids
are said to be miscible when they can dissolve in each other. Therefore if they are immiscible, they are
insoluble. When making a cake using a cake mix, you often use 1+ cups of water and 1/3 cup of oil and
then you add this to the mix. If you first mix this together you would see that the mixture is an immiscible
solution. The oil does not dissolve in the water. Since cooking oil is non-polar and the water is polar, they
have different types of intermolecular bonds and they will not make a solution. When making biscuits, on
the other hand, the recipe may call for you to add a little vinegar to the water before adding the liquid
to the dry ingredients. The vinegar will be miscible in water because both vinegar and water are polar
compounds and therefore have the same type of intermolecular bonds and can make a solution.

Saturated and Unsaturated

A saturated solution is one in which a given amount of solvent has dissolved the absolute maximum solute
at that temperature. Let's go back to our table salt and water example from before: now try to dissolve
2 teaspoons of table salt in one cup of water. This is probably the maximum amount of salt that could
dissolve. Try now to dissolve 3 teaspoons, some of the table salt would probably sit on the bottom of the
glass; but at 2 teaspoons, all of the salt is in solution. The solution becomes saturated. If more solute
is added to a saturated solution, the excess solute remains undissolved and simply sits on the bottom of
the cup. If only one (1) teaspoon was placed in the glass, the solution would be said to be unsaturated.
This is because the solvent is holding less than the maximum amount of solute at that temperature. An
unsaturated solution is one that contains less than the maximum amount of solute that is possible in a
given amount of solvent.

Concentrated and Dilute

Solutions can also be said to be dilute or concentrated. A dilute solution is a concentrated solution that has
been, in essence, watered down. Think of the frozen juice containers you buy in the grocery store. What
you have to do is take the frozen juice from inside these containers and usually empty 3 or 4 times the
container size full of water to mix with the juice concentrate and make your container of juice. Therefore,
you are diluting the concentrated juice. When we talk about solute and solvent, the concentrated solution

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has a lot of solute verses the dilute solution that would have a smaller amount of solute. A concentrated
solution is one in which there is a large amount of solute in a given amount of solvent. A dilute solution
is one in which there is a small amount of solute in a given amount of solvent.

Lesson Summary

• Generally speaking, in a solution, a solute is present in the least amount (less than 50% of the
solution) whereas the solvent is present in the greater amount (more than 50% of the solution).

• When a substance can dissolve in another it is said to be soluble; when it cannot, it is said to be
insoluble. For two liquids, when they are soluble in each other the liquids are said to be miscible;
when they are insoluble the liquids are considered immiscible.

• A saturated solution holds the maximum amount of solid at a specific temperature. An unsaturated
solution does not have the maximum amount of solute dissolved at that temperature in a given
amount of solvent. A concentrated solution has a large amount of solute in a given amount of
solvent. A dilute solution has a lesser amount of solute.

Review Questions

1. Distinguish between soluble, insoluble and miscible, immiscible. Use an example in your answer.

2. How can a solution that is concentrated be made more dilute and a dilute be made more concentrated?

3. Vinegar and water will mix together. Therefore two liquids are said to be:

(a) saturated

(b) miscible

(d) immiscible

4. A solution is analyzed and found to contain 90 g of solute in 100 mL of solution. What can be
concluded about this solution?

(a) The solution is concentrated.

(b) The concentration of the solution is 90 g/100 mL of water.

(d) The solution is holding the maximum amount of solute.

5. A solute is defined as:

(a) The substance in a solution present in the least amount.

(b) The substance in a solution that represents less than 50% of the solution.

(d) All of the above.

(e) None of the above.

6. Match the following words with the examples that describe them.

(a) solute - Adding only one can of water to a frozen concentrated juice mix will form this type of
solution.

(b) solvent - Adding eight cans of water to a frozen concentrated juice mix will form this type of
solution.

(d) insoluble - Gasoline and water will have this property.

(e) miscible - The water of a NaOH{aq) solution.

(f) immiscible - When salt is added to water it is said to have this property.

(g) saturated - The copper(II) sulfate crystals in a solution of CuS 0±(aq)

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(h) unsaturated - The maximum amount of silver nitrate that can dissolve in 100 mL of water is

220 g. What term is given to this solution?
(i) concentrated - If 220 g of AgNO% can dissolve in 100 g of water and only 50 g are added, what

type solution is formed?
(j) dilute - Adding calcium hydroxide to water forms a milky white precipitate. What term is given

to calcium hydroxide?

7. Nisi is given two bottles of copper (II) sulfate solutions in her senior high chemistry lab. She is told
that one bottle contains a saturated solution and the other one contains an unsaturated solution.
What can Nisi do to identify the two solutions?

8. Can you have a solution that is saturated and dilute at the same time? Explain.

Vocabulary

solute The substance in a solution present in the least amount.

solvent The substance in a solution present in the greatest amount.

soluble The ability to dissolve in solution.

insoluble The inability to dissolve in solution.

miscible Two liquids having the ability to be soluble in each other.

immiscible Two liquids not having the ability to be soluble in each other.

saturated A solution holding the maximum amount of solution in a given amount of solvent.

unsaturated A solution holding less than the maximum amount of solution in a given amount of solvent.

concentrated A solution where there is a large amount of solute in a given amount of solvent.

dilute A solution where there is a small amount of solute in a given amount of solvent.

15.4 Measuring Concentration

Lesson Objectives

• Define molarity, mass percent, ppm, and molality.

• Calculate molarity, mass percent, ppm, and molality.

• Explain the importance of quantitative measurement in concentration.

Introduction

Although qualitative observations are necessary and have their place in every part of science, including
chemistry, we have seen throughout our study of science that there is a definite need for quantitative
measurements in science. This is particularly true in solution chemistry. We might read in the headlines

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that the amount of mercury found in the fish is up by 0.5ppm and say to ourselves, what does that mean?
Is it important? We read labels in the grocery store that are in weight percent. What does this mean? So
being able to deal with the quantitative side of solutions helps us to move toward a deeper understanding
of solutions, one that involves not only a numerical analysis but a critical analysis as well. Let's explore
some of the different quantitative applications of solution chemistry.

Molarity

Of all the quantitative measures of concentration, molarity is the one used most frequently by chemists.
Molarity is defined as the moles of solute per liter of solution. The symbol given for molarity is M. You
will see molarity units of both moles/liter and M. Chemists also used square brackets [ ] to indicate a
reference to the molarity of a substance. For example, the expression [Ag + ] refers to the molarity of the
silver ion.

It should be noted that when making solutions to a certain molarity, the amount of solvent to be used
cannot be measured. The amount of solvent used will be whatever is necessary to bring the total solution
to the required volume. This amount of solvent cannot be calculated beforehand either by volume or by
mass because it is not known what volume will be required to reach the total volume of solution. Solution
concentrations expressed in molarity are the easiest to calculate with but the most difficult to make in the
lab.

moles of solute
molarity = mols/L

liters of solution

Sample question 1: What is the concentration, in mol/L, where 2.34 mol of NaCl has been dissolved in
500 mL of H 2 0.

Solution:

[NaCl] = 2M m ° 1S = 4.68M
L J 0.500 liters

The concentration of the NaCl solution is 4.68 mol/L

Sample question 2: What would be the mass of 500 mL of a 1.25 mol/L potassium sulfate solution?

Solution:

mol
M = so mols M x L

mols = (1.25 mol/L(0.500L) = 0.625 mol

mols = so mass = (mols) (molar mass)

molar mass

mass = (0.625 mol) (174.3 g/mol) = 109 g
Therefore the mass of K2SO4 that dissolves in 500 mL of H2O to make this solution is 109 g.

Mass Percent

Mass percent, is the number of grams of the solute in the number of grams of solution. Mass percent is a
term frequently used when referring to solid solutions. It has the formula:

mass of solute

percent by mass = X 100

solute mass + solvent mass

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mass solute

percent by mass = X 100

mass solution

Sample question: An alloy is prepared by adding 15 g of zinc to 65 g of copper. What is the mass percent
of zinc?

Solution:

mass of solute

percent by mass = X 100

solute mass + solvent mass

mass Zn 15 g 15 g _,

percent by mass = — x 100 = x 100 = — - x 100 = 19%

mass Zn + mass Cu 15 g + 65 g 80 g

Parts Per Million

Parts per million is another unit for concentration. Obscure in the sense that it is used less frequently
than the others that will be dealt with in this unit. This does not mean that it is any less important
than the others, it just means that for the normal "day to day" conversations that folks have around the
solution chemistry lab, parts per million might not come up all that often. Parts per million denotes that
there is 1 milligram of solute for every kilogram of solvent. Usually you hear about parts per million when
governments are talking to us about drinking water or poisons in fish and other food products. It is used
most frequently when dealing with environmental issues when you pick up the newspaper or common news
magazines around the house. To calculate parts per million, the following formula is used.

mass of solute R

ppm = t — rr^~ x 10

mass oi solution

Sample question: Mercury levels in fish have often been at the forefront of the news for people who love to
eat fresh fish. Salmon, for instance, contains O.Olppm compared to shark which contains 0.99 ppm. In the
United States, canned tuna is the most popular selling fish and has a mercury level of 0.12 ppm according
to the FDA statistics. If one were to consume 1.00kg of canned tuna over a certain time period, how much
mercury would be consumed?

Solution

mass of solute p

ppm = t — rr^~ x 10

mass oi solution

mass of solute

(mass of solution) (ppm)

1 x 10 6
(1000. g)(0.12) 10 _ n _ 4

mass of solute = 7 . = 1.2 x 10

1 x 10 6

Molality

Molality is one further way to measure concentration of a solution. It is calculated by dividing the moles
of solute by the kilograms of solvent. Molality has the symbol, m.

. ,. , . moles solute

molality (m) =

kg of solvent

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An interesting sidebar: molality is used for specific topics in solution chemistry as molarity as a method for
measuring concentration. So why learn both? Well, oddly enough it all boils down to temperature ... pardon
the pun! Molarity, if you recall, is moles of solute per volume of solution and the volume is temperature
dependent. As the temperature rises, the molarity of the solution will actually decrease slightly because the
volume will increase slightly. Molality does not involve volume and mass is not temperature dependent.
Thus, there is a slight advantage to using molality over molarity when temperatures move away from
standard conditions. Yet, still, the choice of majority falls to using molarity.

Sample question: Calculate the molality of a solution of hydrochloric acid where 12. 5g of hydrochloric acid
has been dissolved in 115 g of water.

Solution:

12 5 e:

mol HC1 = —^ — = 0.343 mol

36.46 g/mol

TT ^, moles solute 0.343 mol

molality HC1 = = — = 2.98 m

kg solvent 0.115 kg

Lesson Summary

• Molarity is the moles of solute per liter of solution. Molarity normally uses the symbol M. Mass
percent is the number of grams of the solute in the number of grams of solution, of course multiplied
by 100.

• Parts per million means that there is 1 milligram of solute for every kilogram of solvent. Therefore
it is the mass of solute per mass of solution multiplied by 1 million.

• Molality is calculated by dividing the moles of solute by the kilograms of solvent. It is less common
than molarity but more accurate because of its lack of dependence on temperature.

Review Questions

1. Calculate the mass percent of silver when a silver /nickel solution is made with 34.5 g of silver and
72.3 g of nickel.

2. What would be the ppm of silver for the data presented in question 1?

3. Why is it a good idea to learn mass percent when molarity and molality are the most commonly used
concentration measures?

4. Most times when news reports indicate the amount of lead or mercury found in foods, they use the
concentration measures of ppb (parts per billion) or ppm (parts per million). Why use these over
the others we have learned?

5. What is the molarity of a solution prepared by dissolving 2.5 g of LiNO% in sufficient water to make
60 mL of solution?

(a) 0.036 mol/L

(b) 0.041 mol/L

(d) 0.060 mol/L

6. A solution is known to have a concentration of 325 ppm. What is the mass of the solute dissolved in
1.50 kg of solvent?

(a) 0.32 mg

(b) 0.49 mg

(d) 488 mg

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7. Calculate the molality of a solution of copper (II) sulfate where 11.25 g of the crystals has been
dissolved in 325 g of water.

(a) 0.0346 m

(b) 0.0705 m

(d) None of the above

8. What is the mass of magnesium chloride present in a 250 g solution found to be 21.4% MgCl<p.

(a) 21.4 g

(b) 53.5 g

(d) 250 g

9. What is the concentration of each of the following solutions in mol/L.

(a) 3.50 g of potassium chromate dissolved in 100 mL of water

(b) 50.0 g of magnesium nitrate dissolved in 250 mL of water.

10. Find the mass of aluminum nitrate required to produce 750 g of a 1.5 molal solution.

11. The Dead Sea contains approximately 332 grams of salt per kilogram of seawater. Assume this salt
is all NaCl. Given that the density of the Dead Sea water is approximately 1.20 g/mL, calculate:

(a) the mass percent of NaCl.

(b) the mole fraction of NaCl.

Further Reading / Supplemental Links

Website with lessons, worksheets, and quizzes on various high school chemistry topics.

• Lesson 6-4 is on Molarity.

• http : //www . f ordhamprep . org/gcurran/sho/sho/lessons/lesson31 . htm

Vocabulary

molarity A concentration unit measuring the moles of solute per liter of solution.

mass percent A concentration unit measuring the mass of solute per mass of solution. This unit is
presented as a percent.

weight percent Another name for mass percent.

parts per million (ppm) A concentration unit measuring the mass of solute per mass of solution multi-
plied by 1 million.

molality A concentration unit measuring the moles of solute per kilograms of solutions.
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15.5 Solubility Graphs

Lesson Objectives

• Define solubility.

• Read and report data from solubility graphs.

• Read and report saturation points from a solubility graph.

Introduction

In an earlier chapter we have discussed solubility as it applied to data analysis. Solubility graphs are
an excellent way of organizing and displaying data for interpretation. In this lesson, we will explore this
concept in learning how to read and analyze a solubility graph in order to extract the relevant data. In
the everyday world, solutions play a key role in our lives from the foods we eat (through proper mixing),
the solutions we prepare to clean homes, and also are essential in the laboratory.

Reading Solubility From a Graph

Solubility, as we already know, is the amount of solute that will dissolve in a given amount of solvent
at a particular temperature. The latter part of this statement is significant since, for many solutes, the
solubility will increase as the temperature is increased. There are exceptions, of course, just as there are
exceptions to every rule. Sodium chloride (table salt), for example, will dissolve to the extent of about 36 g
in 100 g of water at 25°C and there is little change as the temperature increases. The solubility of cesium
sulfate actually decreases as temperature increases. But a vast majority of ionic solids that are solutes do
increase their solubility with temperature.

Think of the solutes you would be dissolving in your own environment. If you make a cup of instant coffee
you most likely boil the water first to dissolve more of the coffee crystals. If you are making hot chocolate,
you would also use boiling water; hot tap water would not do the trick quite so well. Making a bowl of
"Cup of Noodles" simply does not taste quite as good when you do not have the noodles heated in the
boiling water because the noodles cannot dissolve all of the spices to the flavor that is expected. Think
also of a can of cola. This is a solution of a gas, CO2, in a liquid. When you have a can of soda at room
temperature there is more gas above the surface of the liquid in the can than if you have a can of soda
cooled to an icy cold temperature. This is because the solubility of the gas in solution decreases as the
temperature is increased and the gas moves up to the air space in the can above the liquid. Think of this
before you open up the next warm pop.

To display the different solubilities at different temperatures, a solubility graph is drawn to show the data
in a more coherent manner. Having a solubility graph allows us to read the data about a particular solute
or to compare solutes at a particular temperature quickly and easily. Let's look at a typical solubility
graph and see how it works.

(See Figure 15.2).

What kind of information does this graph tell us? You can see that three of the four solids increase
solubility with increasing temperature, NaCl only slightly, and KNO3 solid increases substantially with
increasing temperature. In addition to general trends in the solubility of a substance, you can also get
detailed facts from a solubility graph. For example, we can see that at 30°C, 95 g of sodium nitrate,
NaNOs, will dissolve but at 60°C, 120 g will dissolve in 100 g of H2O. At these same two temperatures,
only 50 g of NCI2SO4 and 113 g, of potassium nitrate, KNO3, will dissolve in 100 g of H2O.

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Figure 15.2: A solubility graph for some common compounds.

Sample question 1: Answer the following questions using Figure 15.2.

(a) How much sodium nitrate will dissolve at 30°C?

(b) What solid is more soluble at 60°C?

(d) At what temperature will 60g of sodium sulfate dissolve in 100 g of water?
Solution:

(a) Looking at the solubility graph (see Figure ??), you draw a line up (vertically) from 30°C until it
hits the NaNOz line. Following this, carry the line over (horizontally) to find the amount of NaNO^ that
dissolves.

Therefore approximately 95 g of NaNO^ will dissolve in 100 g of water at 30° C.

(b) The highest line at 60° C is the green line (NaNO^), therefore it is the most soluble at 60° C.

(d) Looking at the solubility graph (see Figure 15.4), you draw a line over (horizontally) from 60 g until
it hits the Na2SO^ line. Following this, carry the line down (vertically) to find the temperature at which
60 g of Nci2SOa will dissolve.

Therefore 60 g of Na2SO^ will dissolve in 100 g of water at 50°C

Reading Saturated Solutions From a Graph

Look at the solubility graph that shows more common ionic compounds. The lines on the solubility curves
represent the amounts that dissolve in the given amount of solvent at a specific temperature. Look at
the line for NH3. According to the graph that NH3 is the only substance of this group that decreases
in solubility as the temperature is increased. We can also see that the most soluble substance at room
temperature (25°C) is NH3 because it is the line highest up on the graph at 25°C The highest point on

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Figure 15.3: Reading a solubility graph.

Figure 15.4: Reading a solubility graph.

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the solubility curve is at approximately y = 92. We can say then that the most soluble substance at 0°C
is ammonia with a solubility of approximately 92 g in 100 g of water.

(See Figure 15.5).

Solubility «i

i 1 islu t« 1C.-C!

Figure 15.5: A solubility graph for common ions and ammonia.

All of this information can be obtained from reading the solubility graph. What other information can you
obtain from a solubility graph? You could do a number of different types of calculations. For example,
what if you were doing an experiment in the lab (at room temperature) and needed a saturated solution
of potassium chloride dissolved in 35 g of water. How much KCl would you need?

At 25°C (room temperature), approximately 35 g of KCl will dissolve in 100. g of water. For 35 g of water:

Proportion

x g KCl 35 g KCl

35 g H 2 100. g H 2
x = 12g KCl

Another type of problem that can be answered using a solubility graph is to determine if you have saturated
solutions or not. For example, you have a solution of potassium chlorate that you know is 76 g dissolved in
250 g of water. You want to know if this solution is saturated or unsaturated when your solution is being
heated at 80° C.

Looking at the solubility graph, at 80°C, 44 g of KC10 3 will dissolve in 100 g of H 2 0. Therefore we can
use the same type of equation as used previously to determine how much would dissolve in 250 g of H 2 0.

Proportion

x g KC10 3 44 g KC10 3

250. g H 2

100. g H 2
HO.g KC10 3

Since it is possible to dissolve 110 g of KC10 3 in 250 g of H 2 and our solution only has 76 g dissolved in
250 g of H 2 0, the solution is unsaturated.

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Lesson Summary

• Solubility is the amount of solute that will dissolve in a given amount of solvent at a specific temper-
ature. A solubility graph is drawn to display the solubility at different temperatures. It is the mass
of the solute/100 g of H^O versus temperature in °C.

• From reading a solubility graph, one can determine the mass of solute that can dissolve at specific
temperatures and also compare solubility's of different substances at specific temperatures.

Review Questions

1. Using the graph below, determine:

(a) How much ammonia will dissolve at 30° C?

(b) What solid is more soluble at 50° C?

(d) At what temperature will 50 g of ammonium chloride dissolve in 100 g of water?

Soluburcr

<;'' ■■■■'■' I !'"■:

NHjCI

^-f-~"""'"T KCICtj

L^-^T~ -^-s^"* KCI

J— — "j""" ^f

\j^

T*mp4rMUr*^C|

2. Why are solubility graphs useful?

3. Define solubility and solubility graph.

4. How many grams of NaCl are in 450 g of water at 30°C if the solubility is 39.8 g per 100 g of water?

(a) 8.84 g

(b) 39.8 g

(d) 179 g

5. How many moles of ammonium chloride are in 225 g of water at 40°C if the solubility is 45.8 g per
100 g of water?

(a) 0.86 mol

(b) 1.92 mol

(d) 103 mol

6. How many moles of potassium chloride are in 500 g of water at 80°C if the solubility is 51.3 g per
100 g of water?

(a) 0.140 mol

(b) 0.688 mol

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(d) 10.3 mol

7. Plot the following data on a solubility graph and then answer the questions below.

(a) Which substance is the most soluble at 50°C?

(b) Which substance is the least soluble at 90°C?

(d) How many grams of NH 4 C10 4 would dissolve in 250 mL at 30° C?

(e) At what temperature will 20 g potassium sulfate dissolve in 100 g of water?

Table 15.3:

Temp (°C) g NH 4 Br/100 g H 2 g NH 4 ClO 4 /100 g H 2 g NaClO 3 /100 g H 2

60.0 13.0 80.0

20 75.5 23.5 98.0

40 92.0 36.8 118.0

60 107.8 51.5 143.0

80 126.0 67.9 172.0

100 146.0 87.0 207.0

8. Plot the following data on a solubility graph and then answer the questions below.

(a) Which substance is the most soluble at 50°C?

(b) Which substance is the least soluble at 90°C?

(d) At what temperature will 20 g potassium sulfate dissolve in 100 g of water?

Table 15.4:

Temp (°C) g NaCl/ 100 g H 2 g K 2 SO 4 /l00 g H 2 g CuSO 4 /100 g H 2

35.7 7.4 14.3

20 36.0 11.1 20.7

40 36.5 14.8 28.7

60 37.3 18.2 40.0

80 38.1 21.4 56.0

100 39.2 24.1 80.0

Vocabulary

solubility The amount of solute that will dissolve in a given amount of solvent at a particular tempera-
ture.

solubility graph A solubility graph is drawn to display the solubility at different temperatures. It is
the mass of the solute/100 g of H 2 versus temperature in °C.

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15.6 Factors Affecting Solubility

Lesson Objectives

Describe the factors that affect solid solubility.
Describe the factors that affect gas solubility.
Describe how pressure can affect solubility.

Introduction

Solubility, as we have learned, is the maximum amount of a substance that will dissolve in a given amount
of solvent at a specific temperature. There are two direct factors that affect solubility, temperature and
pressure. The effects of temperature on solubility can be found for both solids and gases but pressure
effects are only related to the solubility of gases. Surface area, while not affecting how much of a solute
will be dissolved, is a factor in how quickly or slowly the saturation point will be reached. In this section,
we will explore all three of these factors and how they affect the solubility of solids and gases.

The Effect of Temperature on Solubility

As we learned in the solubility graphs from the previous section, temperature has a direct effect on solubility.
For the majority of ionic solids, increasing the temperature increases how quickly that solution can be made.
As the temperature increases, the particles of the solid move faster which increases the chances that they
will interact with more of the solvent particles. This results in increasing the rate at which a solution
occurs.

However, temperature can also increase the amount of solute that can be dissolved in a solvent. Generally
speaking, as the temperature is increased, more solute particles will be dissolved. Do take note that this
is a generalization but one with which you might already be familiar. For instance, when you add table
sugar to water, a solution is quite easily made. When you heat that solution and keep adding sugar, you
find that large amounts of sugar can be added as the temperature keeps rising. This is how candy is made.
The reason this occurs is that as the temperature increases, the intermolecular bonds can be more easily
broken which allows more of the solute particles to be attracted to the solvent particles.

There are other examples, though, when increasing the temperature has very little effect on how much
solute can be dissolved. Table salt is a good example: you can dissolve just about the same amount of
table salt in ice water as you can in boiling water.

Chemistry often has exceptions to the rules. For all gases, as the temperature increases, the solubility
decreases. Using the kinetic molecular theory to explain this phenomenon, as the temperature increases,
the gas molecules move faster and are then able to escape from the liquid. Their solubility then decreases.

Look at the graph below, ammonia gas, NH3, shows a sharp decline in solubility as the temperature
increases whereas all of the ionic solids show an increase in solubility as the temperature increases.

Solubility graph for ionic solids and NH3.

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-j

NHjCI

N. ^_^ x -~-*~"^

KClOb

(S soliri*-l«g

KCI

~-~~\

"~— -|-^_

! ^^

» 40 K « tM
TfeHMHtflirVfCi

Remember from earlier chapters, that endothermic changes absorb energy and exothermic changes release
energy. When gases dissolve in water, it is usually with an exothermic change. Notice that in Figure 1, the
NH$(g) solubility decreases with temperature and as we look at the equation for the dissolution of NH^(g),
we see that it is exothermic.

NH 3(g) -* NH 3(aq) + energy

A graph for the solubility of oxygen gas, O2, would be very similar to the one for NH^(g); in other
words, oxygen gas would decrease in solubility as the temperature rises. Or, conversely, the colder the
temperature, the greater amount of 02(g) would be dissolved. This is an important environmental concept
in understanding the effect of the increase in the temperature in the world's oceans and its effect on the
oceanic life.

The Effect of Pressure

The second factor, pressure, affects the solubility of a gas in a liquid but never a solid dissolving in a liquid.
When pressure is applied to a gas that is above the surface of a solvent, the gas will move into the solvent
and occupy some of the spaces between the particles of the solvent. A good example is carbonated soda.
Pressure is applied to force the CO2 molecules into the soda. The opposite is also true. When the gas
pressure is decreased, the solubility of that gas is decreased. One example you can think about is when you
open a can of soda pop, or any carbonated beverage, for that matter. When you hear the "pop" as a can
or bottle of soda pop is opened, the pressure in the soda is lowered and the gas immediately starts leaving
the solution. The carbon dioxide stored in the soda is released and you see the fizzing on the surface of
the liquid.

This gas pressure factor is expressed in Henry's Law. Henry's Law states that at a given temperature
the solubility of a gas in a liquid is proportional to the pressure of the gas above the liquid. An example of
Henry's Law occurs in SCUBA diving. As a person dives into deep water, the pressure increases and more
gases are dissolved into the blood. While ascending from a deep water dive, the diver needs to return to
the surface of the water at a very slow rate since all of the dissolved gases have to come out of the blood
very slowly. If for some reason a person ascends too quickly, a medical emergency may occur since the
gases will come out of the blood too quickly. This is called having the "bends".

Increased Surface Area Increases Rate of Dissolving

Another factor that affects the rate of solubility is the surface area of a solid. If we were to increase the
surface area of a solid then it would have been broken into smaller pieces. We would do this to increase

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how quickly the solute would dissolve in solution. If you were to dissolve sugar in water, a sugar cube will
dissolve slower than an equal amount of tiny pieces of sugar crystals. The combined surface area of all of
the sugar crystals have a much greater surface area than the one sugar cube and therefore will have more
contact with the water molecules. This allows the sugar crystals to dissolve much more quickly. Try it on
your own and see! What other examples can you think of?

If you were working in a lab, you might be asked to make a solution of copper(II) sulfate. Copper(II) sulfate
comes in several forms: one being a large beautiful blue crystal and the other ground-up blue crystals.
When you set equal amounts of each of these is test tubes filled with 10 mL of water, you will notice after
5 minutes that more of the ground-up crystal will have dissolved (and the solution will be a darker blue)
than the test tube with the large crystal. Or, you can take two samples of the ground-up crystal and put
into separate test tubes. This time stopper one of the test tubes and carefully shake it while you let the
other test tube simple sit still. Again you are increasing the surface area by increasing the how much of the
ground-up crystal will come in contact with the water. The result will still be the same as before: the test
tube with the greater surface area will go into solution at a faster rate. Even though maximum solubility
is achieved more quickly with greater surface area, the concentration of the solute at maximum solubility
will be exactly the same. (See Figures 15.6 and 15.7).

Figure 15.6: Crystalline copper (II) sulfate.

Figure 15.7: Powdered copper (II) sulfate.

Lesson Summary

Temperature affects the solubility of both gases and solids. With solids, generally the solubility
increases with increasing temperature. With gases, the solubility tends to decrease with increasing
temperature.

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• Pressure only affects the solubility of gases. Henry's Law states that at a given temperature the
solubility of a gas in a liquid is proportional to the pressure of that gas.

• Increasing the surface area increases the rate of solubility of a solid because a larger number of
molecules of the greater surface area have contact with the solvent.

Review Questions

1. What are the factors that affect solubility?

2. What is Henry's Law?

3. Is it ever possible to have ionic solids decrease solubility with increasing temperature?

4. What factor would affect the solubility of sodium sulfate?

(a) i, ii, and iii

(b) i and ii

(d) ii and iii

i. temperature
ii. pressure
iii. surface area

5. What factor would affect the solubility of methane?

(a) i, ii, and iii

(b) i and ii

(d) ii and iii

i. temperature
ii. pressure
iii. surface area

6. If you crush a cube a sugar before putting it in your cup of coffee, how have you affected its solubility?

(a) Crushing it has really no affect on solubility because we have not heated it at all.

(b) Crushing it has increased the surface area so it speeds up the dissolving process but doesn't
change maximum solubility.

(d) Crushing it has increased the surface area so it increases the maximum solubility.

7. Why do people add chlorine to their swimming pools on a hot day?

8. Explain why crushed table salt at room temperature dissolves faster than rock salt.

9. Under which of the following sets of conditions would the solubility of CO2C?) be lowest? The pressure
given is the pressure of C 02(g) above the solution.

(a) 5.0 atm and 75° C

(b) 1.0 atm and 75° C

(d) 1.0 atm and 25°C

(e) 3.0 atm and 25°C

10. An aqueous solution of KCl is heat from 15°C to 85°C Which of the following properties of the
solution remain the same?

(a) i only

(b) iii only

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(d) ii and iii only

(e) i, ii, and iii

i. molality
ii. molarity
iii. density

Vocabulary

Henry's Law At a given temperature the solubility of a gas in a liquid is proportional to the pressure
of that gas.

15.7 Colligative Properties

Lesson Objectives

• Describe vapor pressure lowering.

• Define boiling point elevation and freezing point depression.

• Describe what happens to the boiling points and freezing points when a solute is added to a solvent.

• Describe the importance of the van't Hoff factor.

• Calculate the boiling point elevation for electrolyte and non-electrolyte solutions.

• Calculate the freezing point depression for electrolyte and non-electrolyte solutions.

Pure solvents and solutions of solutes in solvents vary in their boiling points and freezing points. People
who live in colder climates have seen the trucks put salt on the roads when snow or ice is forecast. Why
do they do that? When planes fly in cold weather, the planes need to be de-iced before liftoff. Why is that
done? In this lesson you will learn why these events occur. You will also learn to calculate exactly how
much of an effect a specific solute can have on the boiling point or freezing point of a solution.

Vapor Pressure Lowering

In the chapter on liquids, you studied the concept of vapor pressure for liquids. An enclosed liquid will
reach a vapor pressure equilibrium with its vapor in the space above the liquid and this vapor pressure
depends on the temperature of the liquid. This vapor pressure equilibrium is reached when the rate of
evaporation and the rate of condensation become equal. Raising the temperature increases the rate of
evaporation and therefore, increases the vapor pressure of the liquid. When the temperature of the liquid
becomes high enough for the vapor pressure to equal the surrounding pressure, the liquid will boil. When
the surrounding pressure is 1.00 atm, the boiling point is called the "normal" boiling point.

A pure liquid solvent, at normal atmospheric pressure, cannot have its temperature raised above the normal
boiling point because at the normal boiling point, all of the liquid will vaporize before the temperature
will increase.

Adding a solute to a solvent lowers the vapor pressure of the solvent. There are two suggested explanations
for why the addition of a solute lowers the vapor pressure of a solution. Since they seem equally valid,
both will be presented here.

Remember that only the molecules on the surface of a liquid are able to evaporate. (See Figure 15.8).

In a pure solvent, all the molecules at the surface are solvent molecules. Therefore, the entire surface area
is available for evaporation and the forces to be overcome are the attractive forces between the solvent

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Pure Solvent Solution

Figure 15.8: Pure solvent particles versus intermixed solute particles.

molecules. One of the explanations says that in a solution, some of the surface molecules are solute
molecules and since these solute molecules take up some of the surface area, less surface area is available
for evaporation. Therefore, the rate of evaporation of the solvent will be lower and so the vapor pressure
will be lower at the same temperature. The other explanation says that the attractive forces between the
solvent molecules and the solute molecules are greater than the attractive forces between solvent molecules
and therefore, the solvent molecules will not evaporate at as high a rate. Once again vapor pressure will
be lowered. Both explanations start with the same premises and end with the same result so there doesn't
seem to be a reason to choose between them.

The amount of vapor pressure lowering is related to the molal concentration of the solute. Suppose a
pure solvent has a vapor pressure of 20. mm of Hg at a given temperature. Suppose further that sufficient
non-electrolyte solvent is dissolved in the solvent to make a 0.20 m solution and that the vapor pressure
of the solution at the same temperature is 18 mm of Hg. Then, if enough solute is dissolved to make the
solution 0.40 m, the vapor pressure of the more concentration solution will be 16 mm of Hg. There is a
direct mathematical relationship between the molality of the solution and the vapor pressure lowering.

Experiments with this phenomenon demonstrate that as long as the solute is a non-electrolyte, the effect
is the same regardless of what solute is used. The effect is related only to the number of particles of solute,
not the chemical composition of the solute. That is the definition of the term colligative properties;
properties that are due only to the number of particles in solution and not related to the chemical properties
of the solute. Any polar, non-ionic solid used to make a 0.20 m solution would cause the exact same vapor
pressure lowering.

Boiling Point Elevation

The boiling point of a liquid occurs when the vapor pressure above the surface of the liquid equals the
surrounding pressure. We know that water boils at 100°C at 1 atm but a solution of salt water does not.
When table salt is added to water the resulting solution has a higher boiling point because the salt causes
the vapor pressure of the solvent to lower. The ions form an attraction with the solvent particles which then
prevent the water molecules from going into the vapor phase. Pure water at 100°C has a vapor pressure
of 1.00 atm and since the surrounding pressure is 1.00 atm, the pure water boils at this temperature. If
some quantity of table salt is added to the water, the vapor pressure of the solution at 100°C might be
only 0.98 atm. Therefore, the salt water solution will not boil at 100°C In order to cause the salt water
solution to boil, the temperature must be raised above 100°C in order to get the vapor pressure up to
1.00 atm and allow the solution to boil. This is true for any solute added to a solvent; the boiling point
of the solution will be higher than the boiling point of the pure solvent (without the solute). The boiling
point elevation is the difference in the boiling points of the pure solvent and the solution.

The boiling point elevation due to the presence of a solute is also a colligative property. That is, the
amount of change in the boiling point is related to number of particles of solute in a solution and is not
related to chemical composition of the solute. A 0.20 m solution of table salt and a 0.20 m solution of
hydrochloric acid would have the same effect on the boiling point.

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Freezing Point Depression

The effect of adding a solute to a solvent has the opposite effect on the freezing points of a solution as it
does on the boiling point. A solution will have a lower freezing point than a pure solvent. The freezing
point is the temperature at which the liquid changes to a solid. At a given temperature, if a substance is
added to a solvent (such as water), the solute-solvent interactions prevent the solvent from going into the
solid phase. The solute-solvent interactions require the temperature to decrease further in order to solidify
the solution. A common example is found when salt is used on icy roadways. Here the salt is put on the
roads so that the water on the roads will not freeze at the normal 0°C but at a lower temperature, under
ideal conditions to as low as — 9°C. The de-icing of planes is another common example of freezing point
depression in action. A number of solutions are used but commonly a solution such as ethylene glycol, or
a less toxic monopropylene glycol, is used to de-ice an aircraft. The aircrafts are sprayed with the solution
in weather conditions suspected of dropping below the freezing point. The freezing point depression is
the difference in the freezing points of the solution from the pure solvent.

Mathematics of Boiling Point and Freezing Point Changes

The boiling point of a solution is higher than the boiling point of a pure solvent and the freezing point of
a solution is lower than the freezing point of a pure solvent. However, the amount to which the boiling
point increases or the freezing point decreases depends on the amount solute that is added to the solvent.
A mathematical equation is used to calculate the difference in difference of the boiling point elevation or
the freezing point depression. The boiling point elevation is the difference between the boiling points of
the pure solvent and the solution. Remember the solution has a higher boiling point so to find the boiling
point elevation you would subtract the boiling point of the solvent from the boiling point of the solution.
For example, the boiling point of pure water at 1.0 atm is 100. °C while the boiling point of a 2% salt
water solution is about 102°C Therefore, the boiling point elevation would be 2°C The freezing point
depression is the difference in the freezing points of the solution from the pure solvent. Remember that
the solution has a lower freezing point so the freezing point depression is found by subtracting the freezing
point of the solution from the freezing point of the pure solvent.

Both the boiling point elevation and the freezing point depression are related to the molality of the solutions.
Looking at the formulas for the boiling point elevation and freezing point depression, we can see similarities
between the two.

ATb = Kbin

ATf = Kfm

where:

ATt, = T(pure solution)-r(pure solvent)
ATf = T(pure solvent)-r(pure solution)
Kb = boiling point elevation constant
Kf = freezing point depression constant
m = molality of the solution
m = molality of the solution

The boiling point constants and freezing point constants are different for every solvent and are determined
experimentally in the lab. You can find these constants for hundreds of solvents listed in data reference
publications for chemistry and physics.

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Sample question 1: Antifreeze is used in automobile radiators to keep the coolant from freezing. In
geographical areas where winter temperatures go below the freezing point of water, using pure water as
the coolant could allow the water to freeze. Since water expands when it freezes, freezing coolant could
crack engine blocks, radiators, and coolant lines. The main component in antifreeze is ethylene glycol,
C2Hi{OH)2- If the addition of an unknown amount of ethylene glycol to 150 g of water dropped the
freezing point of the solution by — 1.86°C, what mass of ethylene glycol was used? The freezing point
constant, Kf for water is -1.86° C/m.

Solution:

ATf = Kfin

At f -1.86°C

m = — - = — - — = 1.00 m

K f -1.86°C/m

mols solute

kg solvent

moles solute = (molality) ( kg solvent) = (1.00 mol/kg) (0.150 kg) = 0.150 mol
mass C2Hi{OH)2 = ( mols)( molar mass) = (0.150 mol)(62.1 g/mol)
mass C 2 H 4 {OH) 2 = 9.3 g

Therefore 9.3 g of ethylene glycol would have been added to the 150. g of water to lower the freezing point
by 1.86°C.

One factor that must be considered when using these formulas is to first determine whether or not the
solution is an electrolyte or a non-electrolyte.

Recall that an electrolyte solution is one where the solute will separate into ions and is thus able to conduct
electricity (hence the term electrolyte). If, for example, you have a solution of sodium chloride, the sodium
chloride separates into sodium ions and chloride ions. A non-electrolyte solution does not separate into
ions and therefore does not conduct electricity. For example a sugar /water solution stays as sugar + water
with the sugar molecules staying as molecules.

Remember that colligative properties are due to the number of solute particles in the solution. Adding
10 molecules of sugar to a solvent will produce 10 solute particles in the solution. When the solute is an
electrolyte, such as NaCl however, adding 10 molecules of solute to the solution will produce 20 ions (solute
particles) in the solution. Therefore, adding enough NaCl solute to a solvent to produce a 0.20 m solution
will have twice the effect of adding enough sugar to a solvent to produce a 0.20 m solution. It is the
number of solute particles in the solution that control the colligative properties.

To add this "electrolyte" factor into the formulas above, we use a concept called the van't Hoff factor,
symbolized by the letter i. The van't Hoff factor is the number of particles that the solute will dissociate
into upon mixing with the solvent. For example, sodium chloride, NaCl, will dissociate into two ions so
the van't Hoff factor for NaCl is i = 2, for lithium nitrate, LiNOz, i = 2, and for calcium chloride, CaCh,
i = 3.

We can now rewrite our colligative properties formulas and include the van't Hoff factor.

AT i, — iKbtn
ATf = iKfin

where:

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ATb = T(pure solvent )-T (pure solution)
ATf = T(pure solvent )-T (pure solution)
i = van't Hoff factor
Kb = boiling point elevation constant
Kf = freezing point depression constant
m = molality of the solution

Why can we use this formula for both electrolyte and non-electrolyte solutions? Since the van't Hoff factor
for non-electrolytes is always 1 (because they do not dissociate), i is always equal to 1.

Sample question 2: A solution of 10.0 g of sodium chloride is added to 100.0 g of water in an attempt to
elevate the boiling point. What is the boiling point of the solution?

Solution:

ATb — iKbtn

mols NaCl = f™ = 1 °-° / S - = 0.171 mols
molar mass 58.5 g/mol

, , mols solute 0.171 mol „ „ _, . , „, .

molality = = = 1.71 m For NaCLj = 2 (NaCl -> Na + + CV)

kg solvent 0.100 kg v '

K b ( water) = 0.52° C/m

ATb = iKb m

AT h = (2)(0.52° C/m)(1.71 m)

AT,, = 1.78°C
^(solution) = 7^ (pure solvent) + ATb
resolution) = 100°C + 1.78°C = 101.78°C.

Therefore the boiling point of the solution of 10 g of NaCl in 100 g of water is 102°C

Lesson Summary

• Boiling points of solutions are higher that the boiling points of the pure solvents. Freezing points
of solutions are lower than the freezing points of the pure solvents. Boiling point elevation is the
difference between the boiling points of the pure solvent and the solution. The boiling point elevation
can be calculated using the formula ATb = Kb m, where ATb is the boiling point elevation, Kb is the
boiling point elevation constant, and m is molality.

• Freezing point depression is the difference between the freezing points of the solution and the pure
solvent. The freezing point depression can be calculated using the formula ATf = Kf m, where ATf
is the freezing point depression, Kf is the freezing point depression constant, and m is molality.

• For electrolyte solutions, the van't Hoff factor is added to account for the number of ions that the
solute will dissociate into in solution. For non-electrolyte solutions, the van't Hoff factor = 1. The
equations change to account for this factor (ATb = Kb m becomes ATb = 1Kb m and ATf = Kf m
becomes ATf = iKf m, where i is the number of particles each solute molecule produces in solution.

Review Questions

1. What must be measured in order to determine the freezing point depression?

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2. From a list of solutions with similar molalities, how could you quickly determine which would have
the highest boiling point?

3. Why would table salt not be a good solution to use when deicing a plane?

4. When a solute is added to a solution:

(a) i and hi are true

(b) i and iv are true

(d) ii and iv are true

i. the boiling point increases.

ii. the boiling point decreases,

iii. the freezing point increases,

iv. the freezing point decreases.

5. If 25.0 g of sucrose (C12H22O11) is added to 500. g of water, the boiling point is increased by what
amount? (K b ( water) = 0.52° C/m)

(a) 0.076°

(b) 0.025°

(d) None of these

6. The solubility of seawater (an aqueous solution of NaCl) is approximately 0.50 m. Calculate the
freezing point of seawater. (Kf ( water) = 1.86° C/m)

(a) -0.93°

(b) 0.93°

(d) -1.86°

7. Determine which of the following solutions would have the lowest freezing point.

(a) 15 g of ammonium nitrate in 100. g of water.

(b) 50. g of glucose in 100. g of water.

8. A 135.0 g sample of an unknown nonelectrolyte compound is dissolved in 725 g of water. The
boiling point of the resulting solution was found to be 106.02°C What is the molecular weight of the
unknown compound?

9. What is the van't Hoff factor for each of the following:

(a) MgCl 2

(b) Ammonium sulfate

(d) Potassium chloride

(e) KCH 3 COO

10. Calcium chloride is known to melt ice faster than sodium chloride but is not used on roads because
the salt itself attracts water. If 15 g of CaCli was added to 250 g of water, what would be the effect
on the freezing point of the solution? (Kf( water) = -1.86° C/m)

Vocabulary

boiling point elevation The difference in the boiling points of the pure solvent from the solution.

freezing point depression The difference in the freezing points of the solution from the pure solvent.
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Van't Hoff factor The number of particles that the solute will dissociate into upon mixing with the
solvent

15.8 Colloids

Lesson Objectives

• Define colloids and suspensions.

• Compare solutions, colloids, and suspensions.

• Characterize solutions as suspensions, colloids, or solutions.

• Name some common examples of colloids.

Introduction

In this final section regarding solution chemistry, we will take a look at colloids and suspensions; two
different types of mixtures that are not classified as true solutions. Following this, a look at the different
types of colloids in terms of the solute/solvent combinations will be examined to see how we have many
examples in our everyday lives. Do you know that the majority of our encounters with mixtures are actually
colloids? Think of that the next time you drink a glass of milk, put whipping cream on your dessert, look
up at the clouds, or add some butter to your toast!

Comparison of Suspensions, Colloids and Solutions

We learned early in the chapter that a solution is a mixture of substances in such a way that the final
product has the same composition throughout. Remember the example of vinegar that is 5%, by mass,
acetic acid in water. This clear liquid is a solution since light easily passes through it and it never separates.

On the other hand, colloids are mixtures in which the size of the particles is between 1 X 10 3 pm and
1 X 10 6 pm. In meters, these sizes translate to 1 X 10~ 9 m to 1 X 1CT 6 m; a small grain of sand has a diameter
of 2 x 1CT 5 m. A common example for a colloid is milk. One way to tell that milk is a colloid is to test
it using the Tyndall Effect. The Tyndall effect involves shining a light through the mixture: when the
light is not allowed to pass through the mixture, that is, the light is scattered, the mixture is considered
a colloid. This is why milk appears "cloudy" - or what we think of as "milky". In contrast, when light is
passed through a true solution, the particles in solution are so small [atoms, ions, small molecules] that
they do not obstruct the light. However, when light is passed through a colloid, since the particles are
larger, they will act as an obstruction to the light and the light is scattered. However, these particles,
while able to scatter light, are still small enough so that they do not settle out of solution.

In contrast, suspensions are mixtures where the particles settle to the bottom of the container which
means that the particles in a suspension are large enough so that gravity pulls them out of solution. With
suspensions, filtration is usually able to be used to separate the excess particles from the solution. A
common example of a suspension is muddy water. If you had a beaker of water and added a handful of
fine dirt, even if you stirred it (making a colloid type solution), letting it stand, dirt would settle to the
bottom.

Many Common Products Are Colloids

It is amazing just how common colloids are to us in our everyday lives. In our earlier lessons we discussed
solutes and solvents and what types of solutions formed as a result. In this final section of this final lesson,

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let's do the same process. Looking at Table 15.5, we see that some common colloid products that are
formed when different phase solutes and solvents are mixed.

Table 15.5:

Solid Solvent

Liquid Solvent

Gas Solvent

Solid Solute Ruby, brass, steel Butter, cheese, mus- Marshmallow

tard, Jell-0
Liquid Solute Paint, milk of magnesia Milk, Mayonnaise, Whipping Cream, Shav-

Creams (i.e. face ing Cream

creams)
Gas Solute Airborne viruses, car Fog, Smoke, Clouds, Aerosol Sprays

exhaust, smoke

Lesson Summary

Colloids are mixtures in which the size of the particles is between 1 x 10 3 pm and 1 x 10 6 pm. Suspensions
are mixtures in which the particles are large enough so that they settle to the bottom of the container and
can be filtered using filter paper. The Tyndall effect involves shining a light through the mixture, if the
light scatters, the mixture is a colloid or a suspension. Common examples of colloids include milk, butter,
Jell-O, and clouds.

Review Questions

1. Distinguish between a solution, a colloid, and a suspension.

2. What is one true way to tell you have a colloid solution?

3. Why do you think there is no example of a gas - gas colloid?

4. Which is an example of a colloid?

(a) air

(b) brass

(d) none of these

5. Which is not an example of a colloid?

(a) human body

(b) mayonnaise

(d) cloud

6. The biggest difference between a colloid and a suspension is that:

(a) In colloids, the solute is permanently dissolved in the solvent.

(b) In colloids the particles eventually settle to the bottom.

(d) None of these are correct

7. Karen was working in the lab with an unknown solution. She noticed that there was no precipitate
in the bottom of the beaker even after it had been on the lab bench for several days. She tested it
with a light and saw that light scattered as it passed through the solution. Karen concluded that the
liquid was what type of a mixture?

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(a) colloid

(b) suspension

(d) heterogeneous

8. What are two good common examples of colloids?

Vocabulary

colloid Mixtures where the size of the particles is between 1 x 10 3 pm and 1 x 10 6 pm (i.e., milk).

suspension Mixtures where the particles settles to the bottom of the container and can be separated by
filtration.

Tyndall Effect Involves shining a light through the mixture, if the light scatters, the mixture is a colloid.

15.9 Separating Mixtures

Lesson Objectives

• The student will describe differences between the physical properties of pure substances and solutions.

• The student will list and describe methods of separation for mixtures.

• The student will explain the principles involved in chromatographic separation.

• The student will identify the mobile and stationary phases in a chromatography design.

• Given appropriate data, the student will calculate Rf values.

Introduction

Mixtures occur very commonly in chemistry. When a new substance is synthesized, for example, the new
substance usually must be separated from various side-products, catalysts, and any excess reagent still
present. When a substance must be isolated from a natural biological source, the substance of interest is
generally found in a very complex mixture with many other substances, all of which must be removed.
Chemists have developed a series of standard methods for the separation of mixtures. (The separation of
mixtures into their constituent substances defines an entire sub-field of chemistry referred to as separation
science.

Differing Solubilities

Mixtures of solids may often be separated on the basis of differing solubilities of the solids. If one
of the components of the mixture is soluble in water while the other components are insoluble in water,
the water-soluble component can be removed from the mixture by dissolving the mixture in water and
filtering the mixture through filter paper. The component dissolved in water will pass through the filter
while the undissolved solids will be caught in the filter. The solubility of substances is greatly influenced
by temperature. By controlling the temperature at which solution occurs or at which the filtration is
performed, it may be possible to separate the components. Most commonly, a sample is added to water
and heated to boiling. The hot sample is then filtered to remove completely insoluble substances. The
sample is then cooled to room temperature or below which causes crystallization of those substances whose

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solubilities are very temperature dependent. These crystals can then be separated by another filtration and
the filtrate (the stuff that went through the filter) will then contain only those substances whose solubilities
are not as temperature dependent.

Distillation

Homogeneous solutions are most commonly separated by distillation. (See Figure 15.9). In general,
distillation involves heating a liquid to its boiling point, then collecting, cooling, and condensing the
vapor produced into a separate container. For example, salt water can be desalinated by boiling off and
condensing the water.

In solutions of non-volatile solid solutes in liquid solvent, when the solution is boiled, only the solvent boils
off, all the solid remains in the solution. The vapor passing off is pure solvent. As the solvent passes off
and the all the solute remains behind, the same amount of solute is now dissolved in less solvent and so
the concentration increases. This results in the boiling point of the solution increasing. As a solution boils,
increased temperature is necessary to keep the solution boiling because its boiling point has increased. This
is a quick method of determining if a liquid is a pure substance or a solution; start it boiling and if it
continues to boil at the same temperature, it is a pure substance whereas if its boiling point increases, it is
a solution.

For a mixture of liquids in which several components of the mixture are likely to be volatile (easily vapor-
ized), the separation is not easy. If the components of the mixture differ reasonably in their boiling points,
it may be possible to separate the mixture simply by monitoring the temperature of the vapor produced
as the mixture is heated. Liquid components of a mixture will each boil in turn as the temperature is
gradually increased, with a sharp rise in the temperature of the vapor being distilled indicating when a
new component of the mixture has begun to boil. By changing the receiving flask at the correct moment,
a separation can be accomplished. This process is known as fractional distillation.

Figure 15.9: Distillation Apparatus

Chromatography

The word chromatography means color- writing. The name was chosen around 1900 when the method was
first used to separate colored components from plant leaves. Chromatography in its various forms is perhaps

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the most important known method of chemical analysis of mixtures. Paper and thin-layer chromatography
are simple techniques that can be used to separate mixtures into the individual components. The methods
are very similar in operation and principle and differ primarily in the medium used.

Paper chromatography uses ordinary filter paper as the medium upon which the mixture to be separated is
applied. Thin-layer chromatography (abbreviated TLC) uses a thin coating of aluminum oxide or silicagel
on a glass microscope slide or plastic sheet to which the mixture is applied. A single drop of the unknown
mixture to be separated is applied about half an inch from the end of a strip of filter paper or TCL slide.
The filter paper or TLC slide is then placed in a shallow layer of solvent in a jar or beaker. Since filter
paper or the TLC slide coating is permeable to liquids, the solvent begins rising up the paper by capillary
action.

As the solvent rises to the level of the mixture spot, various effects can occur, depending on the constituents
of the spot. Those components of the spot that are completely soluble in the solvent will be swept along
with the solvent front as it continues to rise. Those components that are not at all soluble will be left
behind at the original location of the spot. Most components of the mixture will move up the paper or
slide at an intermediate speed somewhat less than the solvent front speed but not remaining at the original
spot either. In this way, the original spot of mixture is spread out into a series of spots or bands, with each
spot representing one single component of the mixture. The separation of a mixture by chromatography
is not only a function of the solubility in the solvent used. The filter paper or TLC coating consists of
molecules that may interact with the molecules of mixture as they are carried up the medium. The primary
interaction between the mixture components and the medium is due to the polarity of the components and
that of the medium. Each component of the mixture is likely to interact with the medium to a different
extent thus slowing the components of the mixture differentially depending on the level of interaction. (See
Figure 15.10).

Solvent front

Intermediate Spots

Origbal Spot

Line of Origb

Figure 15.10: Paper chromatography strip

A quantitative basis is added to the chromatography analysis using a mathematical function call the
retention factor. The retention factor, Rf, is defined as

distance travelled by spot
distance travelled by solvent front

Rf is the ratio of the distance a substance moves up the stationary phase to the distance the solvent have
moved. The retention factor depends on what solvent is used and on the specific composition of the filter
paper or slide coating used. The Rf value is characteristic of a substance when the same solvent and the
same type of stationary phase is used. Therefore, a set of known substances can be analyzed at the same

449

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time under the same conditions and if the same Rf value is produced for a known and unknown, it is a
step toward identifying the unknown.

In the case shown at right, the Rf for the green spot is

2.7 cm

5.7 cm

0.47

and for the yellow spot

1.8 cm

R f = — = 0.32

5.7 cm

2.7 cm

1.8 cm

Paper chromatography and TLC are only two examples of many different chromatographic methods. Mix-
tures of gases are commonly separated by gas chromatography. In this method, a mixture of liquids are
vaporized and passed through a long tube of solid absorbent material. A carrier gas, usually helium, is
used to carry the mixture of gases through the tube. As with paper chromatography, the components of
the mixture will have different solubilities and different attractions for the solid absorbent. Separation of
the components occurs as the mixture moves through the tube. The individual components exit the tube
one by one and can be collected.

Another form of chromatography is column chromatography. In this form, a vertical column is filled with
solid absorbent, the mixture is poured in at the top, and a carrier solvent is added. As the mixture flows
down the column, the components are separated, again, by differing solubilities in the carrier solvent and
different absorbencies to the solid packing. As the liquid drips out the bottom of the column, components
of the solution will exit at different times and can be collected.

Lesson Summary

• Mixtures of solids may be separated by differing solubilities of the solids.

• Components of a solution composed of a non- volatile solid solute and a liquid solvent can be separated
by distillation.

• Mixtures of liquids with reasonably different boiling points can also be separated by distillation.

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450

• Solutions with several components can be separated by paper or thin-layer chromatography.

• Gas chromatography and column chromatography are also used to separate the components of a
solution.

Review Questions

1. In a paper chromatography experiment to separate the various pigments in chlorophyll, a mixture of
water and ethanol was used as the solvent. What is the stationary phase in this separation?

2. Do you think that paper chromatography or TLC would be useful for separating a very large quantity
of a mixture? Explain why or why not.

3. If the mobile phase in a chromatographic experiment moved 15.0 cm and one of the compounds in
the mixture moved 12.7 cm, what is the Rf value for this compound?

4. If the stationary phase in a paper chromatography experiment was very polar and the solvent was
moderately polar, would the polar components in the mixture be closer to the bottom of the paper
or toward the top of the paper?

Vocabulary

distillation The evaporation and subsequent collection of a liquid by condensation as a means of purifi-
cation.

fractional distillation This is a special type of distillation used to separate a mixture of liquids using
their differences in boiling points.

chromatography Any of various techniques for the separation of complex mixtures that rely on the
differential affinities of substances for a mobile solvent and a stationary medium through which they
pass.

Image Sources

(2
(3
(4
(5

(10

Powdered copper (II) sulfate.. Public Domain.

Therese Forsythe. A solubility graph for common ions and ammonia.. CC-BY-SA.
CK-12 Foundation. Reading a solubility graph.. CC-BY-SA.

Crystalline copper (II) sulfate.. Public Domain.
H Padleckas. Distillation Apparatus. GNU Free Documentation License.
CK-12 Foundation. Reading a solubility graph.. CC-BY-SA.
Richard Parsons. The hydration of ions in a polar solvent.. CC-BY-SA.
Richard Parsons. Paper chromatography strip. CC-BY-SA.
Richard Parsons. . CC-BY-SA.
CK-12 Foundation. . CC-BY-SA.

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Chapter 16

Ions in Solution

16.1 Ionic Solutions

Lesson Objectives

• Describe electrostatic attraction.

• Explain how ionic solids attract water molecules when they dissolve in water.

• Explain the difference between physical changes and chemical changes.

• Define electrolyte solutions and be able to identify electrolytes.

Introduction

Ionic solids are a particular type of substance. They form when metal ions combine with nonmetal ions,
which come about through the transfer of electrons. Because of this transfer and the distinct charges that
result when the metals and nonmetal ions form, ionic solids have properties that are unique to themselves.
In this lesson we will begin to look at the forces that exist within the ionic solids and what happens when
these solids dissolve in water. There are some new terms in this unit and a few reviewed terms that you
have seen in previous units that will become clearer in this chapter.

Ions in Solids are Held With Electrostatic Attraction

In an earlier chapter you learned that metals form positive ions by losing electrons and nonmetals form
negative ions by gaining electrons. When solids form from a metal atom donating an electron (thus forming
a positive cation) to a nonmetal atom (thus forming a negative anion) , the ions in the solid are held together
by the attraction of these oppositely charges particles. The attraction of oppositely charged particles is
called electrostatic attraction. For example:

Na + ionization energy — » Na + + e~

I + e~ — » I + energy of electron affinity

Na + + I — » Nal + energy

(Nal is held together by electrostatic attraction)

In the world of chemical bonding, electrostatic attraction is quite strong and therefore compounds with

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this type of bonding have high melting points and boiling points. Take a look at Table 16.1 and see how
the melting points and boiling points change for ionic solids.

Table 16.1:

Compound

Melting Point (°C)

Boiling Point (°C)

LiCl

NaCl

KCl

MgCl 2

CaCh

613
801

772
714

772

1360
1413
1500
1412
1935

All of the compounds in the table are bonded due to the attraction of the oppositely charged ions. Look
at the next examples to see lithium, sodium, and potassium chlorides.

Li — > Li + + e-
Cl +e- — > Cl-
Li + +C|-— ► LiCl

Na — > Na + + e
CI + e - — » CI -
Na + +CI- — > NaCl

NaCl

(s)

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K — ► K + + e
CI +e~ — *CI~
K + + CI - — ► KCI

KCI

(s)

To review from a previous chapter, ionic compounds do not form molecules. The empirical formula repre-
sents the lowest whole number ratio of the ions involved in the compound. When we interact with these
substance in our environment, we detect that they are crystalline structures. As shown in the diagrams,
ionic compounds are crystals that are held together by electrostatic attraction.

Sample question 1: Which compounds would contain an electrostatic attraction as bonds between ions?

(a) MgCh

(b) Al 2 3

Solution:

(a) and (b) would contain electrostatic attraction because they are both ionic but (c) is not. It does not
form bonds by the transfer of electrons but rather by sharing electrons and therefore does not have ions
for electrostatic attraction.

Ions Are Attracted to the Polar Water Molecule

Since ionic compounds can dissolve in polar solutions, specifically water, we can extend this concept to say
that ions themselves are attracted to the water molecules because the ions of the ionic solid are attracted
to the polar water molecule. When you dissolve table salt in a cup of water, the table salt dissociates into
sodium ions and chloride ions (Equation 1).

NaCl(s) — > Na + ( aq -j + CI r aq \ (Equation 1)

The sodium ions then get attracted to the negative ends of the water molecule and the chloride ions get
attracted to the positive end of the water molecule. The process of water molecules attaching to ions is
called hydration. Look at the Figure ??.

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Sodium Chloride
Crystal

Na +

Na + CI- Na +
Na +

Hydrated Na + ion

Na + Cl-
Na +
Na + CI- Na +

Hydrated Ch ion

8 « *

CI- Na + Cl|

The same is true for any ionic compound dissolving in water. The ionic compound will separate into the
positive and negative ions and the positive ion will be attracted to the negative end of the water molecules
(oxygen) while the negative ion will be attracted to the positive end of the water molecules (hydrogen) .

Dissociation is a Physical Change

Matter can go through both chemical changes and physical changes. Chemical changes are ones that
occur with the chemical bonding and a new substance or substances are formed. For example, if you add
a piece of lead to a solution of silver nitrate, silver precipitates out and lead nitrate is formed (Equation

2).

Pb (s) + 2 AgN0 3{aq) -> 2 Ag (s) + Pb(N0 3 ) 2(aq) (Equation 2)

Of importance to us in this unit are physical changes. Physical changes are those that occur in the
physical state of the substance but do not affect the identity of the substance. For example, grinding a
sugar cube is a physical change or dissolving table salt in water; the sugar is still sugar and the table salt
is still table salt. Look at the reaction for the dissolving of copper(II) sulfate in water (Equation 4).

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CuSO^s) — » Cu + (aq) +SO4 fa) (Equation 4)

We begin with copper(II) sulfate as a solid that contains Cu 2+ ions and SO 2 ' ions and it produces Cu 2+
ions and S0 2 ~ ions in solution. We dissolved a sample of CuSO^ in a beaker of water. If we look at this
change, we may first think it is a chemical change because we might think that we have new substances
being formed. But realistically, if we boiled off the water containing the dissolved copper (II) sulfate ions,
the CuSO^(s) would remain in the beaker. In other words, we can regain our original starting material by
a simple physical change, that is, evaporation. Think of the crystal of copper(II) sulfate as a combination
of the Cu 2+ ions and S0 2 ~ ions that have taken position and then when dissolved in water, lose their
position. When the water is evaporated, they regain their position. Then the dissolving of an ionic solid
seems more like a physical change.

Sample question: Which of the following are physical changes?

(a) burning paper

(b) melting wax

(a) burning paper is a chemical change (new substances are produced)

(b) melting wax and (c) evaporating water are physical changes as no new substances are produced, (both
involve a change in state)

Solutions of Electrolytes Can Conduct Electricity

When we started this lesson we said that ionic compounds are held together by electrostatic attraction.
Arrhenius in the late 1800s first thought that when these ionic compounds dissolved in water, they separated
from each other into ions and classified them as electrolytes. Since they could conduct electricity in water
solution, they are considered electrolytes. According to Arrhenius (and current theory - pardon the pun!),
the ions in solution provided the charged particles needed to conduct electricity. Look at the equation
below for the dissociation of NaCl.

NaCl (s) -> Na + (aq) + Cl~ {aq)

The sodium and chlorine ions are present in the crystal. But once the solid NaCl is added to the water,
it dissolves, which means that the ions move away from their crystalline structure and are now dispersed
throughout the water molecules. If two electrodes were to be immersed into a solution of NaCl(aq),
the Na + {aq) ions would move toward one electrode and the Cl'(aq) ions would move toward the second
electrode. This movement of the ions allows the electric current to flow through the solution. Therefore,
NaCl{aq) will behave as an electrolyte and conduct electricity because of the presence of Na + (aq) and
Cl~{aq) ions. The more ions that are present in solution when the salt dissolves, the stronger the electrolyte
solution is.

Sample question: Which of the following will form electrolyte solutions and conduct electricity?

(a) CaF2 (aq)

(b) CqHi 2 O g (aq)

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Solution:

(a) and (c) are solutions that contain positive cations and negative anions that would separate when
dissolved in water. Since ions are separated in solution, they are electrolytes and will conduct electricity.

CaF 2 ( s ) ~ * Ca 2+ {aq) + 2 F~(aq)
Calcium fluoride
KOH(s) -> K + {aq) + OH~(aq)
Potassium hydroxide

(b) Is not an ionic compound but a covalent compound. This means that when it dissolves in water it
stays together as a molecule and is a non-electrolyte

C 6 H 12 6 (s) -» C 6 H 12 6 (aq)

Glucose separate glucose molecules

Lesson Summary

• Electrostatic attraction describes the bonding that occurs between the ions of ionic solids. Because of
the strong electrostatic attraction in ionic solids, ionic compounds tend to have high melting points
and boiling points.

• Ionic compounds dissolve in polar solvents, especially water. This occurs when the positive cation
from the ionic solid is attracted to the negative end of the water molecule (oxygen) and the negative
anion of the ionic solid is attracted to the positive end of the water molecule (hydrogen) .

• Electrolyte solutions are ones in which free-moving charged particles can conduct an electrical current.

Review Questions

1. Write the reactions for the dissolving of the following.

(a) NaOH( s)

(b) LiOH (s)

(d) NH A Cl {s)

(e) MgCl 2{s)

2. Which of the following represent physical changes? Explain.

(a) explosion of TNT

(b) dissolving KCI

(d) souring milk

3. Which compound contains electrostatic forces?

(a) natural gas

(b) table salt

(d) sugar

4. Which of the following is a physical change?

(a) rotting wood

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(b) rising of bread dough

(d) molding cheese

5. Which of the following is not a physical change?

(a) melting iron

(b) pumping gas

(d) reaction of magnesium chloride with water

6. Which compound is considered to be an electrolyte when dissolved in water?

(a) HN0 3

(b) Ci2#22011

(d) CH A

7. Which compound is not considered to be an electrolyte?

(a) AgCl

(b) PbSOi

(d) HC10 3

8. Janet is given three solutions. She is to determine if the solutions are electrolytes or not but is not
told what the solutions are. She makes the following observations. What can you conclude from her
observations and what help can you offer Janet to determine if the solutions are indeed electrolytes?

Solution 1
Solution 2
Solution 3

Clear

Blue but transparent

Clear

Further Reading / Supplemental Links

• http://en.wikipedia.org/wiki

Vocabulary

electrostatic attraction When solids form from a metal atom donating an electron (thus forming a
positive cation) to a non-metal (thus forming a negative anion) the two ions in the solid are held
together by the attraction of oppositely charged particles.

chemical changes Changes that occur with the chemical bonding where a new substance is formed.

physical changes Changes that occur in the physical structure but do not occur at the molecular level.

electrolyte solutions Solutions that contain ions that are able to conduct electricity.

16.2 Covalent Compounds in Solution

Lesson Objectives

• Understand and describe intermolecular bonds.
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• Understand why molecules stay together when dissolving in solvents.

• Understand and define non-electrolytes.

Introduction

Unlike their ionic counterparts, covalent compounds have a different type of attraction occurring between
the solute and solvent molecules. As we work through this lesson, think about the periodic table and
remember that covalent compounds are made up of only nonmetals. This is the foundation for the reasoning
behind the way covalent compounds dissolve.

Molecules Held Together by Intermolecular Forces of Attraction

Unlike ionic bonds that transfer electrons, covalent compounds share electrons to complete an outer electron
configuration. As a result there are no distinct charges associated with the atoms in covalent compounds.
Look at the formula for CO2. Carbon has 4 valence electrons and oxygen has 6 valence electrons.

C : ls 2 2s 2 2p 2
O : ls 2 2s 2 2p 4

Carbon will share four electrons with other atoms and oxygen will share two electrons with other atoms.
In CO2 , each oxygen atom shares two of its electrons with carbon and the carbon shares two of its electrons
with each oxygen atom. Look at the figure below.

:o::c:;o:

• ♦

This sharing of valence electrons represents covalent bonding. However, this sharing of electrons may not
be equal sharing and in the case of carbon and oxygen, it is not equal sharing. An electronegativity table
lists the electronegativities for elements in the periodic table. Elements with a greater electronegativity
have a stronger attraction for shared electrons. Therefore they can pull the electrons closer to themselves
and away from the element that has the more positive electronegativity. For carbon, the electronegativity
value is 2.5, and for oxygen it is 3.5.

The reason that oxygen has a higher electronegativity is due to its larger nuclear charge and smaller
diameter. While both carbon and oxygen have the same shielding of their nuclei, that is, Is 2 , oxygen has
eight protons in its nucleus while carbon only has six. As a result oxygen will pull any shared electrons closer
to it. This pulling of electrons is what is being measured in electronegativity. The result in this molecule
is that the electrons are pulled closer to oxygen than carbon and the resultant structure is represented
below.

Polar bonds in a carbon dioxide molecule (but not a polar molecule). (Source: Richard Parsons. CC-BY-

SA)

The bonding that occurs within the molecule, as we know, involves intramolecular forces. In this molecule,

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while the sharing of electrons inside the molecule is unequal and therefore polar, the overall result produces
a non-polar molecule because the shifting of the shared electrons toward the oxygen atoms are in equal
but opposite directions and there is no dipole moment on the molecule.

When we dissolve a nonpolar covalent compound, such as CO2, in a non-polar solvent, such as benzene,
the attraction between the molecules of CO2 and CqHq would be intermolecular forces of attraction.
The intermolecular attractions that occur between all non-polar substances are London-dispersion forces.
This similarity of intermolecular forces is what allows these two substances to form a solution.

Molecules Separate When Dissolved in Solution

When we studied how ionic solids dissolve, we said that as they dissolve in solution, these solids separate
into ions. More specifically, ionic solids separate into their positive ions and negative ions in solution. The
same is not true for molecular compounds. Molecular compounds are held together with covalent bonds
meaning they share electrons. When they share electrons, their intramolecular bonds do not easily break
apart, thus the molecules stay together even in solution. For example, when you dissolve a spoonful of
sugar into a glass of water, the intermolecular bonds are broken but not the intramolecular bonds.
You can write the following equation for the dissolution of sugar in water.

Ci2#220n(.y) -» C 12 H220u(a q ) (Equation 1)

Notice how the molecules of sugar are now separated by water molecules. In other words, sugar's in-
termolecular bonds are broken but since the molecule has not broken apart, this tells us that the in-
tramolecular bonds have not broken. Since sugar is a polar compound, its intermolecular bonds are
dipole-dipole including some hydrogen bonding. The same is true for water and it is this similarity of
intermolecular bonding that allows the sugar and water to form a solution. Also be sure to take note that
the sugar did not separate into carbon ions, hydrogen ions and oxygen ions but stayed together as a sugar
molecule when dissolved in water. This is characteristic of covalent compounds or compounds formed
between non-metals.

Sample question: Which compounds will dissolve in solution to separate into ions?

(a) LiF

(b) P 2 F 5

LiF will separate into ions when dissolved in solution: LiF{aq) — » Li + {aq) + F~{aq)
P2F5 and C2//5O// are both covalent and will stay as molecules in a solution.

Non-Electrolytes

Remember that solutions of electrolytes conduct electricity and the conduction is the result of ions moving
through a solution. With covalent compounds, there are no ions moving around in solution, therefore they
are classified as non-electrolytes. Non- electrolytes are solutions that do not conduct electricity. If you
were to connect a conductivity meter to these solutions, there would be no reading or a light would not
turn on if the wires were placed in a solution containing a non-electrolyte.

Electrical conduction by an electrolyte. (Source: Richard Parsons. CC-BY-SA)
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Solution of Electrolyte Solution of Non-Electrolyte

A conductivity apparatus is an incomplete electrical circuit that contains a source of electricity and a light
bulb or meter that will show when current is flowing through the circuit. The ends of the incomplete
circuit are prongs that can be lowered into a solution. If the prongs are lowered into a solution containing
a sufficient number of ions, the circuit will be completed by the solution, current will flow, and the light
bulb will light up. If the prongs are lowered into a solution with on ions or an insufficient number of ions,
not enough current will flow to light the bulb.

Lesson Summary

• Covalent compounds share electrons to obtain a completed outer energy level.

• Covalent compounds do not form ions in solution; they stay together as molecules. For example:

Cl2#220n(. v ) -» Cl2#22<?ll( a? )

• Non-electrolytes are solutions that do not conduct electricity.

Review Questions

1. Describe the intermolecular bonding that would occur between glucose, C%H\20%, and water.

2. Define non- electrolyte and give at least one example.

3. How can you tell by looking at a formula that it is most likely a covalent compound? What does this
tell you about the bonding?

4. Describe how you could tell the difference between an electrolyte and a non-electrolyte solution.

5. Looking at the periodic table, which pair of elements will form a compound that is covalent?

(a) Ca and Br

(b) Fe and O

(d) Co and CI

6. Which of the following compounds will conduct the least amount of electricity if dissolved in water?

(a) KN0 3

(b) BaCl 2

(d) C0 2

7. Steve is given five solutions in the lab to identify. He performs a conductivity test, a solubility (in
water) test, crudely measures the hardness of each substance, and determines the melting point using
a melting point apparatus. Some of the melting points, the teacher tells him are too high or low to
measure using the laboratory melting point apparatus so she gives him the melting point. For the
liquids, he determined the boiling points. He gathers all of his data and puts it into a table. His
teacher gives him the names of the five solutions to match his five unknowns to. Can you help Steve

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match the properties of the unknowns (from the table below) to the solution names (found under the
table)?

Table 16.2:

Unknown Sub-

Conductivity

Solubility (:

in Hardness

Melting

Point

Boiling

Point

stance

water)

(°C)

no (aq)

soluble

semi- brittle

164

yes (aq)

soluble

NA (liquid)

100

yes (aq)

soluble

brittle

«800

no (s)

insoluble

soft

yes (s)

soluble

NA (liquid)

118

List of Unknown names:

Sodium chloride

Naphthalene

Sucrose

Hydrochloric acid (dilute)

Acetic acid

8. Predict the type of bonding that will form between the elements sulfur and bromine. Will this
molecule conduct electricity in water solution?

Further Reading / Supplemental Links

• http://en.wikipedia.org/wiki

Vocabulary

intermolecular bonding The bonding that occurs between molecules.
non-electrolytes Solutions that do not conduct electricity.

16.3 Reactions Between Ions in Solutions

Lesson Objectives

• Use the solubility chart and/or solubility rules to determine if substances are soluble in water.

• Use the solubility chart and/or the solubility rules to determine if precipitates will form.

• Write molecular, ionic, and net ionic equations.

• Understand the use of ionic and net ionic equations.

• Identify spectator ions in ionic equations.

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Introduction

Many ionic compounds are said to be soluble in water while others are said to be insoluble. However, no
ionic compound is completely insoluble in water. Every ionic compound dissociates into its ions to some
extent when placed in water. In fact, the solubility of ionic compounds ranges across a full spectrum from
as little solubility as 1 X 10 -100 moles/liter to 20 moles/liter. Most chemistry textbooks contain, somewhere
in them, a chart or table of the solubility of ionic compounds. Most tables divide the ionic compounds into
two groups, "soluble" or "insoluble". Some books add a third category of "slightly soluble", but this term
often becomes synonymous with insoluble. It is clear that in order to take a large spectrum of solubilities
and put them into two categories, a decision must be made on a dividing line. A particular solubility is
chosen as the dividing line and those compounds whole solubility is below that line are called insoluble and
those above the line are called soluble. It is possible to find, in different textbooks, solubility tables that
disagree as to whether or not a particular substance is soluble. If one table had a dividing line of 0.10 M,
silver acetate would be called insoluble whereas a slightly lower dividing line of 0.075 M would list silver
acetate as soluble. So, keep in mind that sometimes, it is important to determine the actual numerical
solubility of a substance instead of to rely on a chart of soluble and insoluble compounds.

When solutes are dissolved in a solution, the solution is transparent so the dissolved solute particles cannot
be visually detected. If undissolved particles are present in a liquid, they form a cloudy barrier to light
passing through the liquid and hence their presence can be detected visually. Eventually, the un-dissolved
particles will settle to the bottom of the container and then it becomes more apparent that an undissolved
solid is present.

When ionic solutions are mixed together, it is often possible to form an insoluble ionic compound even
though both original compounds were soluble. For example, both silver nitrate and sodium chloride are
soluble compounds. In 1.0 M solution of these substances, the compounds would be completely dissociated.
One solution would contain Ag + and NO^. ions and the other solution would contain Na + and Cl~ ions.
When these two solutions are poured together, all of these ions move around in the solution and come
into contact with each other. If a sodium ion temporarily attaches to either a chloride ion or a nitrate
ion, nothing happens because sodium ions are soluble with both these ions and the ions would simply re-
dissociate. If a silver ion temporarily joins a nitrate ion, nothing happens because silver nitrate is soluble.
But, when a silver ion comes into contact with a chloride ion, the two ions join together permanently
because silver chloride is not soluble. When a silver ion combines with a chloride ion, they form an
insoluble solid particle that will not dissolve. Therefore, when two solutions are mixed, a cloudy, non-
transparent substance forms and eventually this substance will settle to the bottom of the container.
When a non-soluble substance is formed in a solution, it is called a precipitate, and the process is called
precipitation.

Formation of a precipitate. (Source: Richard Parsons. CC-BY-SA)

Solution containing
Ag and NO^ ions.

Solution containing
ft and CrOf" ions.

Mixture containing K + and NO s in solution
and an insoluble solid of Ag2CrO-

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Ions in Solution Can React and Produce Precipitates

Once a solid substance has been separated into its ions, the ions are then available for reactions. When a
compound is in the solid state, the ions are held with electrostatic attractions. It must first be dissolved
in solution to allow ions to move freely. Take for example the reaction between sodium chloride and silver
nitrate. Both of these compounds are available commonly in the solid form. First, you would take both
solids and dissolve them in solution, in this case water. (Equation 1 and Equation 2)

AgNO S ( s) -> Ag+ {aq) + N0 3 {aq)

(Equation 1)

NaCl [s) -> Na + (aq) + CI {aq}

(Equation 2)

Once the solid dissolves to separate into its ions in solution, these ions are available to react together in
the chemical reaction (Equation 3).

AgN0 3(aq) + NaCl [aq) -^ AgCl {s) + NaN0 3{aq)

(Equation 3)

Make note that the AgNO^t aq \ and NaCL aq \ reactants show the ions are in solution. In other words
AgNO?,^) is equivalent to Ag + ( aq j +NC>3~ (aq)-, and NaCl( aq ) is equivalent to Na + ( aq -) +C/~ (aq)- The equation
represented in Equation 3 is a double displacement reaction which means the cations exchange anions in
the reactants to form the products. In the laboratory, a precipitate is formed that we determine is silver
chloride.

The same reactions can be seen when substances undergo ionization. Remember that ionization forms
ions in solution. For example, look at Equation 4 for the ionization of sulfuric acid (chemical change) and
Equation 5 for the dissolving of sodium hydroxide (physical change). Notice how they both end up with
ions in solution.

H 2 S 0^ aq) -> 2 H ( aq) + SO4 (j^

NaOH,

{aq)

Na-

irn)

+ OH

(aq)

(Equation 4)
(Equation 5)

When sulfuric acid reacts with sodium hydroxide we have a double displacement reaction where the cations
exchange anions. Equation 6 shows this reaction.

H 2 S0 4(aq) + 2 NaOH {aq) -* 2 H 2 [L) + Na 2 S0 4(aq)

(Equation 6)

Notice in Equation 6, that liquid water is produced, not a solid. As well, the second product is an aqueous
ionic solution containing the ions Na + ^ and SO^ (aq)- What is essential to these equations and equations
like these is to visualize the ionic aqueous solutions as ions in solution.

Most ionic compounds dissociate in water. (Source: Richard Parsons. CC-BY-SA)

When the text says this

think this

AB(aq)

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464

Using Solubility Charts and/or Solubility Rules

How do you know if a precipitate is produced in a double replacement reaction? Not all reactions of this
type will produce a precipitate. Take for example, if you were to mix a solution of table salt, NaCh s \, and
Epsom salts, MgSO^, in water, you would not get a precipitate. The reaction is seen below.

2 NaCl( aq) + MgS 0^ aq) -> Na 2 S A{aq) + MgCl 2 ( aq ) (Equation 7)

So, how would you know when you would get a precipitate and when not? What scientists use is a set of
solubility rules or, even better, a solubility chart. Table 16.3 represents the solubility chart for the most
common cations and anions found in ionic solids.

Table 16.3: Solubility Chart

C 2 H 3 0^r-

col

cio-

CrOf

NO-

o 2 -

PO\

sof

s 2 -

Ag+

Al 3 +

Ba 2 +

Ca 2 +

Cu 2 +

Fe 2+

Fe 3+

K +

Mg 2+

Mn 2 +

Na +

NH+

Pb 2+

Sn 2+

Sn i+

Sr 2+

Zn 2 +

S = soluble in water, I = insoluble in water, ii = partially soluble in water, * = unknown or does not exist.

Let's now see how we use the solubility chart to determine if two compounds will form a precipitate
when they react. If we had a reaction between sodium bromide and silver nitrate, we know that this is
a reaction between two compounds and therefore is a double replacement reaction. How do we know the
states of the products formed? The reaction is seen below.

NaBr {aq) + AgN0 3{aq) -> NaN0 3{?) + AgBr {?) (Equation 8)

Look at the solubility chart and see if you can predict if the reaction will produce any precipitates. If you
look across the row for sodium ion, all sodium compounds are soluble (S), therefore you can fill in this
part of the equation (See Equation 9)

NaBr {aq) + AgN0 3{aq) -> NaN0 3(aq) + AgBr {?) (Equation 9)

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C,HA'

Br"

co a a

ao 3

CrO + 3

NO/

OH" tf-

?o 4 i

so 4 2

S*^

*B*—

* a

V-

Figure 16.1: Solubility chart being used to determine whether the bromide anion and silver cation combi-
nation will produce a precipitate.

If you look across the row for silver, under bromide ion, you find an I for insoluble. Figure 22.4.1 shows
this row of the solubility chart.

Therefore we can complete the equation (Equation 10).

NaBr (aq) + AgN0 3{aq) -> NaN0 3{aq) + AgBr {s) (Equation 10)

Let's try another one. Take a reaction between ammonium phosphate and lead acetate.

2 (NH 4 ) 3 PO i{aq) + 3 Pb(C 2 H 3 2 ) 2{aq) -> 6 NH 4 C 2 H 3 2{aq) + Pb 3 (P0 4 ) 2{s) (Equation 11)

How did we know that the lead(II) phosphate would precipitate from solution? Why did the ammonium
acetate not precipitate? If we follow the ammonium row in the solubility table across to the acetate column,
we find an "S" at the intersection which indicates this compound is soluble. If we follow the lead row across
to the phosphate column, we find an "I" at the intersection, which indicates insolubility and therefore a
precipitate of this compound will form.

As we said at the beginning, rather than using a solubility chart, some scientists simply use a set of
solubility rules. The rules are as follows:

1. All group 1 metals and ammonium compounds are soluble.

2. All nitrates, chlorates, and bicarbonates are soluble.

3. Halides are soluble except for Ag + , Hg 2 + , and Pb 2+

4. Sulfates are soluble except for Ag + , Ba 2+ , Ca 2+ , Hg 2 2 + , S r 2+ , and Pb 2+

5. Carbonates, chromates, phosphates, and sulfides are insoluble except those from rule #1.

6. Hydroxides are insoluble except for those in rule #1, and Ba 2+ .

It is important to remember that this is a priority set of rules. What this means is that Rule #1 is
first. All group 1 metals and ammonium compounds are always soluble. For example, even though sulfide
compounds are rarely soluble for any cation (rule #5), they will be soluble with group 1 metal ions or with
ammonium ions (rule #1). It also does not matter whether you use the set of rules or the solubility chart.

They both provide the same information; the chart is easier to read for some, the rules are easier to
remember for others.

Sample question: Complete the following reactions. Use the solubility table to predict whether precipitates
will form in each of the reactions.

(a) Pb(N0 3 ) 2{aq) + KI {aq) ->

(b) BaCl 2(flq) +Na 2 S0 4(aq) ->
Solution:

(a) Pb(N0 3 ) 2{aq) + 2KI (aq) -> Pbl 2(s) + 2KN0 3{aq)

(b) BaCl 2 ( aq ) +Na 2 SO A ( aq ) -» BaS0 4 ^ + 2NaCl( aq )

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Separating Ions in Solution

When a chemist has a solution that contains two or more ions and he/she wishes to physically separate the
ions, the differences in solubilities can be used. Suppose that a chemist has a solution that contains both
Pb 2+ and Zn 2+ ions. If these two ions were dissolved in the solution as nitrates, then the only anion present
is the nitrate ion. If the chemist added some NaCl to the solution, the zinc ions would remain in solution
because ZnCl^ is soluble but the lead ions and the chloride ions would form an insoluble compound, PbC^,
and form a precipitate. The chemist could then pour this mixture through a piece of filter paper and the
dissolved zinc ions would pass through the filter paper with the solution but the solid PbCli would be
filtered out. Therefore, the chemist would have separated the zinc ions (now in the solution) and the lead
ions (now in the filter paper). See Figure ??.

The process for separating and identifying ions by selective precipitation and filtration is known as gravi-
metric analysis. Analytical chemistry is a sub-discipline of chemistry. The task of analytical chemists is
to identify the substances present in materials and to make quantitative measurements on them. In order
to identify what substances are present, it is often necessary to isolate (separate) the ions from other ions
that might be present. In earlier times, much of this work was done by gravimetric analysis. In modern
times, a large amount of the identification of ions in solution is done by instrumentation.

If you are called upon to determine a process for separating ions from each other, you should look in the
solubility table to determine a reagent that will form a precipitate with one of the ions but not with the
other.

Chemists' knowledge of the interaction of radiation and matter is the basis for analytical methods of
sensitivity and specificity. Signals from Within (http : //www. learner .org/vod/vod_window.html?pid=
802)

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The Reactions Can be Written as Ionic Equations

Remember earlier when we said that we should visualize NaCh aq \ as Na + r aq \ and Cl~ i aq \ and AgN0 3 ( aq \ as
Ag + r aq \ and N0 3 ~t aq \? This is so we could write chemical reactions as formula equations. Equation 12
is a formula equation.

NaCl {aq) + AgN0 3{aq) -> NaN0 3{aq) + AgCl (s) (Equation 12)

The ionic solids discussed in the previous section were written together in a formula form. In an ionic
equation, the separated ions are written in the chemical equation. Let's rewrite Equation 12 as a total
ionic equation which is a better representation of a double replacement reaction.

Na + [aq] + Cr (aq) + Ag + {aq) + N0 3 - {aq) -> Na + {aq) + NOf {aq) + AgCl {s) (Equation 13)

Sample question: Write the ionic equation for each of the following.

(a) BaCl 2 ( aq ) + Na 2 S Atyaci) -> 2 NaCl^ aq ) + BaS 4(i )

(b) 2 K 3 P0 4{aq) + 3 Ca(N0 3 ) 2{aq) -> 6 KN0 3{aq) + Ca 3 (P0 4 ) 2(s)
Solution:

(a) Ba 2+ {aq) + 2 CI [aq) + 2 Na + {aq) + S0 4 2 - (aq) -» 2 A^+ K) + 2 C/- (fl?) + BaS0 4{s)

(b) 6 ^+ H) + 2 POt 3- ^) + 3 Ca 2+ {aq) + 6 MV^) -> 6 K+ {aq) + 6 iV0 3 _ H) + Ca 3 {PO A ) 2(s)

The Essential Information is Contained in a Net Ionic Equation

If you look at Equation 13, what do you notice is the same on both sides of the equation?

Na + {aq) + Cr {aq) + Ag + (aq) + N0 3 ~( aq} -> Na + {aq) + N0 3 ~ \ aq) + AgCl {s) (Equation 13)

Do you see that Na + r aq \ and N0 3 ~ i aq \ appear on both sides of this equation in the same form?

+ Cl(aq) + Ag(aq) +(No,(aq)]— >(Na (aq))+(No,(aq)J+ AgCl(s)

These ions, because they appear on both sides of the equation and in the same form, are called spectator
ions. Think of the phrase "spectator ion". What does it sound like to you? Does it sound like the ion is just
in the container watching the reaction? Well, that's kind of what a spectator ion is doing. A spectator
ion is an ion in the ionic equation that appears in the same form on both sides of the equation indicating
they do not participate in the overall reaction. Therefore Na + (aq) and NO^(aq) are spectator ions for this
reaction. So what is the overall reaction? Well, let's remove the spectator ions and see.

A 8 + { aq ) + cr (aq) -» AgCl {s) (Equation 14)

Without the spectator ions, we see only the ions that are responsible for forming the solid silver chloride.
This equation, represented in Equation 14, is the net ionic equation. The net ionic equation is the overall
reaction that results when spectator ions are removed from the ionic equation. The net ionic equation gives
us all of the essential information we need; what ions we need to form our solid. Really, whether we had
sodium chloride or potassium chloride is irrelevant, what was important was that the chloride ion and the
silver ion were present.

Sample question: Write the net ionic equation for each of the following. Name the spectator ions.
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(a) Ba 2+ {aq) + 2 CI {aq) + 2 Na + (aq) +50 4 2 "( a? ) -» 2 iVa+ (fl9) + 2 C/" W) + flaS0 4( ,)

(b) 6 ^+ H) + 2 P0 4 3 - K) + 3 Ca 2+ {aq) + 6 tf0 3 'te) -» 6 Z+ (as) + 6 tf0 3 '(a,) + Ca 3 (P0 4 ) 2W
Solution:

(a)

Ba 2+ (aq) + 2 C/ - ^) + 2 iVa + (a(? ) + SO* 2- ^ -> 2 Na + {aq) + 2 C/"( fl9 ) + BaS0 4{s)

^ fl (a?) + S0 4 ( ac/ ) — » BaSO^ s ) net ionic equation
Cl~ ( aq ) and Na + ( aq j = spectator ions

(b)

6 K + {aq) + 2 P0 4 3 - M + 3 Ca 2+ {aq) + 6 N0 3 ~ {aq) -» 6 tf+ K) + 6 iVOa"^) + Ca 3 (P0 4 ) 2W

3 Ca 2+ ( aq } + 2 PO^~ t aq \ — > Ca 3 (P0 4 )2(j) net ionic equation
A" + ( a? ) and NO^~ ( aq ) = spectator ions

Lesson Summary

• A solubility chart is a grid showing the possible combinations of cations and anions and their solu-
bilities in water. It is used to determine whether a precipitate is formed in a chemical reaction. The
solubility rules are a list of rules dictating which combinations of cations and anions will be soluble
or insoluble in water.

• A total ionic equation is one in which all of the ions in a reaction are represented. A net ionic
equation is one in which only the ions that produce the precipitate are represented.

Review Questions

1. What is more valuable to use for determining solubility: a solubility chart or a set of solubility rules?

2. If you were told to visualize Cu(NOz)2(aq), what might this mean to you?

3. Use the solubility rules to determine the following solubilities in water. Solubility Rules

(a) All group 1 metals and ammonium compounds are soluble.

(b) All nitrates, chlorates, and bicarbonates are soluble.

(d) Sulfates are soluble except for Ag + , Ba 2+ , Ca 2+ , Hg 2 2+ , Sr 2+ , and Pb 2+

(e) Carbonates, chromates, phosphates, and sulfides are insoluble except those from rule #1.

(f) Hydroxides are insoluble except for those in rule #1, and Ba 2+ .

Which of the following compounds is soluble in water?

(a) PbCl 2

(b) Hg 2 Cl 2

(d) MgC0 3

(e) AgN0 3

(f) MgCl 2

(g) KOH
(h) PbSOi

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4. When only the ions that produce a precipitate are shown for a chemical equation, what type of
reaction exists?

(a) spectator equation

(b) molecular equation

(d) net ionic equation

5. If you wanted to separate a solution of Ag 3+ ( aq ^ from a solution of Hg2 2+ \ a q)-, which of the following
would be the best possible reaction?

(a) add HBr

(b) add HN0 3

(d) none of the above

6. If you wanted to separate a solution of Al + ^ from a solution of Fe 2+ ^, which of the following
would be the best possible reaction?

(a) add HCl

(b) add NaOH

(d) none of the above

7. Complete the following reactions:

(a) Nci2S(aq) + ZnCh(aq) — >

(b) (NH^CO^aq) + CaCl 2 (aq) ->

8. Write the ionic equations for the balanced molecular equations from question 5.

9. Write the net ionic equations for the ionic equations from question 6.
10. Identify the spectator ions for the ionic equations from question 6.

Vocabulary

solubility chart A grid showing the possible combinations of cations and anions and their solubilities
in water.

solubility rules A list of rules dictating which combinations of cations and anions will be soluble or
insoluble in water.

formula equation A chemical equation written such that the aqueous solutions are written in formula
form.

total ionic equation A chemical equation written such that the actual free ions are shown for each
species in aqueous form.

net ionic equation The overall reaction that results when spectator ions are removed from the ionic
equation.

spectator ions The ions in the total ionic equation that appear in the same form on both sides of the
equation indicating they do not participate in the overall reaction.

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Labs and Demonstrations for Ions in Solution
Teacher's Resource Page for Qualitative Ion Testing Lab

Investigation and Experimentation Objectives

In this activity, the student will use self-determined evidence and logically consistent arguments to generate
conclusions. Detailed record keeping of observations is necessary.

Lab Notes

If you have several classes to do the lab, 500 mL of each solution will be adequate. Students have fewer
spills and create less mess with dropper bottles. You will need to refill dropper bottles from larger bottles
after each class. If you purchase a large supply of dropper bottles, the labeled bottles can be stored and
used from year to year. Making and removing labels consumes a large amount of preparation time.

This lab may require more than one day if your lab periods are 50 minutes or less per day. You can also
alter the lab so that your students only test for the cations or for the anions but not both.

Solution Preparation

3.0 M NH^OH Concentrated NH4OH is 14.5 M so 500 mL can be prepared by diluting 103 mL of concen-
trated NH4OH to 500 mL.

3.0 M HNO3 Concentrated nitric acid is 15.6 M so 500 mL can be prepared by diluting 96 mL of concentrated
HNO3 to 500 mL.

0.10 M Ba(NOs)2 Dissolve 13.1 grams of Ba{NOs)2 in sufficient water to make 500 mL of solution. Since
this solution is also used as a testing solution (see below), you can dissolve 26.2 grams in sufficient water
to make 1.0 L of solution and divide the solution to avoid preparing it again.

0.10 M KSCN Dissolve 4.86 grams of KSCN in sufficient water to make 500 mL of solution.

0.10 M AgNOs Dissolve 8.5 grams of AgNOs in sufficient water to make 500 mL of solution. Since this
solution is also used as a testing solution (see below), you can dissolve 17.0 grams in sufficient water to
make 1.0 L of solution and divide the solution to avoid preparing it again.

Unknown salts for students to test, use fiaC/2 or Na2SOi or L/2CO3 (recommended).

For testing solutions:

Barium, Ba 2+ , use 0.1 M Ba{NO^)2', Dissolve 13.1 grams of Ba{NO$)2 in sufficient water to make 500 mL
of solution.

Iron(III), Fe 3+ , use 0.1 M Fe{NO^)y Dissolve 12.1 grams of anhydrous Fe(7V0 3 ) 3 or 20.7 grams of Fe(NOz)%-
9H2O in sufficient water to make 500 mL of solution. Lithium, Li + , use 0.1 M LiNOy Dissolve 3.45 grams of
L1NO3 in sufficient water to make 500 mL of solution. Potassium, K + , use 0.1 M KNO3; Dissolve 5.06 grams
of KNO3 in sufficient water to make 500 mL of solution.

Silver, Ag + , use 0.1 M AgNOy, Dissolve 8.5 grams of AgNO^ in sufficient water to make 500 mL of solution.

Sodium, Na + , use 0.1 M NaNOs; Dissolve 4.25 grams of NaNOz in sufficient water to make 500 mL of
solution.

Carbonate, CO^~, use 0.1 M Na2CO%\ Dissolve 5.30 grams of Na2CO% in sufficient water to make 500 mL
of solution.

Chloride, Cl~, use 0.1 M NaCl] Dissolve 2.93 grams of NaCl in sufficient water to make 500 mL of solution.

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Iodide, / , use 0.1 M Nal; Dissolve 7.50 grams ol Nal in sufficient water to make 500 mL of solution.

Sulfate, SO 2 ', use 0.1 M Na 2 S0 4 ; Dissolve 7.10 grams of Na 2 SOi in sufficient water to make 500 mL of
solution.

For flames tests of Li + , Na + , K + , and Ba 2+ ions, place 25 mL of each of the testing solutions in separate
150 mL (labeled) beakers and stand 15 wooden splints (one for each lab group) in the beakers to soak.

Sometimes, the barrel opening of Bunsen burners are contaminated by previous spills and will produce
colored flames without a testing splint in them. You should check the burners to make sure they do not
produce colored flames when burning. You should also remind students of burner safety procedures . . .
no loose hair that falls past your face, etc.

Answers to Pre-Lab Questions

1. AgN0 3{aq) + Nal {aq)
Yes, precipitate forms, Agl

2. Pb(N0 3 ) 2(aq) + CaCl 2{aq)
Yes, precipitate forms, PbCl 2

3. NH A N0 3{aq) + CaCl 2(aq)
No precipitate forms

4. Sr{N0 3 ) 2{aq) +K 2 SO i{aq)
Yes, precipitate forms, SrSO^

Qualitative Ion Testing Lab

Background:

How are unknown chemicals analyzed? One method is by making comparisons to "known" chemicals. In
this lab activity, ion tests will be performed and observations made for the reactions of four known anions
and six known cations. Then an unknown salt will be identified by analyzing and comparing results to
what is known.

The process of determining the identities of unknown substance is called qualitative analysis. This can be
contrasted to quantitative analysis, which is the process of determining how much of a given component
is present in a sample. Qualitative analysis procedures use physical tests as well as chemical tests. The
physical tests in this lab involve observing colors of solutions and colors produced in flame tests. The
chemical tests in this lab involve chemical reactions, as evidenced by formation of a precipitate, dissolving
of a precipitate to form a complex ion, a color change, or evolution of a gas.

Formation of a Precipitate

An ionic salt is a compound composed of two parts - cations (positively charged ions) and anions (negatively
charged ions). When an ionic salt is dissolved in water, the salt crystal dissociates or separates into its
cations and anions. For example, potassium iodide (KI) dissociates into potassium ions (K + ) and iodide
ions (/"") according to equation 1.

KI (*) -» K tad) + 7 K) Equation 1

Similarly, the ionic salt lead(II) nitrate, Pb(N0 3 ) 2 , dissociates into lead cations, Pb 2+ , and nitrate anions,
N0 3 , according to equation 2.

Pb(N0 3 ) 2(s) -» Pb 2 + } + 2 NO- (aq) Equation 2

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When two ionic salts are mixed together in water, two new combinations of cations and anions are possible.
In some cases, the cation from one salt and the anion from the other salt may combine to form an insoluble
product, which is called a precipitate. For example, if solutions of potassium iodide and lead(II) nitrate
are mixed together, a solid precipitate of lead(II) iodide, Pbl2, forms, as shown in equation 3.

2 K U + 2 7 H) + Pb H) + 2 NO ka t ) - p bh(s) + 2 K+ q) + 2 N0 3{aq) Equation 3

Notice that the potassium cations and the nitrate anions remain dissolved in solution. They did not change
during the reaction and are therefore, referred to as spectator ions. In net ionic equations, spectator ions
are omitted. A net ionic equation is one that includes only the ions participating in the reaction. Thus,
equation 3 can be reduced to equation 4.

Pb \a q ) + 2 7 M "» PM ^) EqUatIOn 4

Dissolving Precipitates through Complex-Ion Formation

A complex ion is a water-soluble, charged species containing a central atom and other molecules bonded to
it. The formation of a complex ion is commonly evidenced by the dissolution of a precipitate. For example,
copper (II) hydroxide, Cu(OH) 2 , is insoluble in water but will dissolve when excess ammonia is added to
it, forming a soluble copper amine complex ion, Cu(NH 3 ) 4 + , according to equation 5.

Cu(OH) 2{s) + 4 NH 3{aq) -» Cu(NH 3 f 4 +{aq) + 2 OH~ {aq) Equation 5

Evolution of a Gas

Certain anions, such as the carbonate ion, CO 2- , and sulfide ion, S 2 ~ , evolve gas when treated with a dilute
strong acid. For example, the reaction of calcium carbonate, CaC0 3 , with nitric acid, HN0 3 , produces
carbon dioxide gas, C0 2 , according to equation 6.

CaC0 3(s) + 2 H+ -> Ca 2 + + C0 2{g) + H 2 (L) Equation 6

Flame Colors

Some metallic salts will display a distinctive color of light when placed in a flame. When the colored light
from any one of these flames is passed through a prism or viewed through a diffraction grating, a portion
of the spectrum is visible, containing only a few colors at specific wavelengths, including the colors in the
original flame. A partial spectrum that contains only discrete lines is called a line spectrum. When heated
in a flame, electrons in the metal absorb energy from the flame and are promoted to excited energy levels.
They emit light as they relax back down to the ground state. Each line in the spectrum represents a
different electronic transition. Since each element has a unique electronic configuration, an element's line
spectrum, and thus its flame color, is unique and can be used for identification.

Pre-Lab Questions

On the last page of this laboratory packet is a listing of solubility rules. Use those rules to determine which
of the following mixtures of solutions would produce a precipitate. Write the formula for the precipitate
where one forms.

1. AgN0 3{aq) +NaI {aq)

2. Pb{N0 3 ) 2{aq) + CaCl 2{aq)

3. NH 4 N0 3{aq) + CaCl 2(aq]

4. Sr{N0 3 ) 2{aq) +K 2 SO i{aq)

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Purpose

In parts I and II of this lab, qualitative tests for four known anions and six known cations will be performed.
Test results will be noted and recorded. In part III, the same tests will be performed on an unknown ionic
salt which contains one of the six possible cations and one of the four possible anions. The cation and
anion that make up the unknown salt will then be identified.

Apparatus and Materials

All solutions should be in drop control dispenser bottles.

Ammonium hydroxide, NH4OH, 3.0 M

Barium nitrate solution, Ba{NOs)2, 0.10 M

Nitric acid solution, HNO3, 3.0 M

Potassium thiocyanate solution, KSCN, 0.10 M

Silver nitrate solution, AgNOj, 0.10 M

Unknown salt solution

Distilled water

Beral-type pipets, labeled

Bunsen burner setup

Reaction plate, 24-well

Sheet of notebook paper

Wooden splints

Beakers, 250 - mL, 15 each

Watchglass, 15 each

Cation Testing Solutions (0.10 M solutions of the nitrates)

• Barium, Ba 2+

. Iron (III), Fe i+

• Lithium, Li +

• Potassium, K +

• Silver, Ag +

• Sodium, Na +

Anion Testing Solutions (0.10 M solutions of the sodium or potassium compounds)

• Carbonate, C0 2 ~

• Chloride, CI

• Iodide, I~

. Sulfate, SOf

Safety Issues

All solutions are irritating to skin, eyes, and mucous membranes, particularly the 3.0 M NH4OH and HNO3
solutions. The silver nitrate solution will turn skin and clothes permanently black. Handle solutions with
care, avoid getting the material on you, and wash your hands carefully before leaving the lab. As with all
labs, do not mix any chemicals other than the ones you are directed to mix.

Procedure

Part I - Anion Testing for Cl~, I , SO 2 ^ and CO\r

Preparing the well plate:

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1. Obtain a 24- well reaction plate and set it on the lab bench with a piece of white paper underneath it.
Label the paper as shown in Data Table 1. Notice that the 24- well reaction plate is divided into 6 columns
(1-6) and 4 rows (A-D).

2. Using a pipet, add 5 drops of the Cl~ anion-testing solution to wells A, B, and C of column 1.

3. Using a pipet, add 5 drops of the I~ anion-testing solution to wells A, B, and C of column 2.

4. Using a pipet, add 5 drops of the SO|~ anion-testing solution to wells A, B, and C of column 3.

5. Using a pipet, add 5 drops of the CO 2 ' anion-testing solution to wells A, B, and C of column 4.
Performing the tests:

Silver Nitrate Test

6. Add 3 drops of 0.10 M AgNO^ to the first four wells across row A. Observe the formation of precipitates
and/or color changes. You may need to remove the paper to see clearly. Record detailed observations in
Data Table 1.

7. Add 5 drops of 3.0 M HNO3 to each of the precipitates from step 6. Gently swirl the well plate to stir.
Observe which precipitates dissolve and which do not. Record observations in Data Table 1.

8. Add 10-12 drops of 3.0 M NH^OH to each of the remaining precipitates from step 6. Gently swirl and
observe which precipitates dissolve and which do not. Record observations in Data Table 1.

Barium Nitrate Test

9. Add 3 drops of 0.10 M Ba{NO^)2 to the first four wells across row B. Observe the formation of precipitates
and/or color changes. You may need to remove the paper to see clearly. Record detailed observations in
Data Table 1.

10. Add 5 drops of 3.0 M HNO3 to each of the precipitates from step 9. Gently swirl the well plate to stir.
Observe which precipitates dissolve and which do not. Record observations in Data Table 1.

11. Add 3 drops of 3.0 M HNO3 to to the first four wells across row C. Make observations, looking for the
strong evolution of gas bubbles. Record observations in Data Table 1.

12. Repeat any tests for which results were unclear in Row D of the well plate. Rinse the plate with plenty
of tap water and then rinse with deionized water to prepare the plate for part II.

Part II - Cation Testing for Li + , Na + , K+, Ag + , Ba 2+ , and Fe 3+ .

Preparing the Well Plate

13. Obtain a 24-well reaction plate and set it on the table with a piece of notebook paper underneath it.
Label the paper as shown in Data Table 2.

14. Using a pipet, add 5 drops of the Li + cation-testing solution to the top well (A) of column 1.

15. Using a pipet, add 5 drops of the Na + cation-testing solution to the top well (A) of column 2

16. Using a pipet, add 5 drops of the K + cation-testing solution to the top well (A) of column 3.

17. Using a pipet, add 5 drops of the Ag + cation-testing solution to the top well (A) of column 4.

18. Using a pipet, add 5 drops of the Ba 2+ cation-testing solution to the top well (A) of column 5.

19. Using a pipet, add 5 drops of the Fe 3+ cation-testing solution to the top well (A) of column 6.
Performing the Tests:

20. Observe each solution and record the color of each solution in Data Table 2 (Row A).
Potassium Thiocyanate Test

21. Add 3 drops of 0.10 M KSCN to each of the six wells across row A. Gently swirl the plate to stir.

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Observe the formation of precipitates and/or color changes. Record detailed observations in Data Table 2.
(Row B)

Flame Tests

Note: Several of the cations may be identified using flame tests. The flame tests will be performed on the
four cations that did not show a reaction in step 21.

22. Set up a Bunsen burner. Adjust the air so the flame color is blue (NO YELLOW) and a distinct inner
blue cone is apparent.

23. Obtain a wooden splint which has been soaking in the Li + cation testing solution for at least 15 minutes.

24. Hold the wooden splint in the flame, flat side down. (Do not touch the top of the burner with the
wooden splint.) The top end of the splint should be placed directly into the inner blue cone. A distinct
color should be apparent. Record the flame color in Data Table 2 (Row C). Do not hold the splint in the
flame too long or the splint will begin to burn.

25. Repeat steps 24 and 25 using the Na + cation-testing solution, then the K + cation-testing solution, and
finally the Ba 2+ cation-testing solution. Be careful not to touch the splints together when gathering them.

26. Rinse the well-plate in the sink with plenty of tap water and make a final rinse with deionized water.
The splint should be discarded in the waste basket.

Part III — Identification of an Unknown Salt

Note: the unknown salt is made up of one of the cations and one of the anions previously tested.

27. Obtain from your teacher a pipet filled with an unknown salt solution and a pre-soaked wooden splint
of the same unknown salt. Be sure to record the unknown identifying letter in the Data Tables.

28. Determine the identity of the cation and the anion that make up the unknown salt. To do this, repeat
the steps in Part I and Part II. Record all observations for anion testing in Data Table 1 and for cation
testing in Data Table 2.

Data for Anion Testing

Record detailed observations inside the circles on the table. Record all colors that form. Record whether
any gases evolve. If any solid precipitates form, use the abbreviation PPT. If no reaction occurs, use the
abbreviation NR.

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Crankm I anion SO 4 anion CO 3 anion Unknown _ Unknown_

12 3 4 5 6

-000000

■000000
000000

000000

Data for Cation Testing

In row A, put the original color of the solution.

In row B, reaction with KSCN.

In row C, flame test color.

Row D is for unknown testing. In column 1, put the original color of the solution. In column 2,

put the results of reaction with KSCN. In column 3, put the flame test color. If you do a second

unknown, use columns 4, 5, and 6.

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CI" anion I" anion S0 4 2 " anion COg 2 " anion Unknown Unknown

12 3 4 5 6

oooooo
■oooooo
oooooo
oooooo

Post-Lab Questions

1. Write the net ionic equation for each precipitation reaction that occurred in Parts I and II. Include the
well identification (Al) as shown in the example below.

Part I - Anion testing

Well Al Ag + + CI -> AgCl( s ) (white ppt)

2. Unknown letter

(a) What cation is present in your unknown?

(b) What anion is present in your unknown?

3. Write the name and formula for your unknown salt.
Solubility Rules

• 1. All group 1 metals and ammonium compounds are soluble.

• 2. All nitrates, chlorates, and bicarbonates are soluble.

• 3. Halides are soluble except for Ag + , Hg 2 + , and Pb 2+ .

• 4. Sulfates are soluble except for Ag + , Ba 2+ , Ca 2+ , Hgl + , Sr 2+ , and Pb 2+ .

• 5. Carbonates, chromates, phosphates, and sulfides are insoluble except those from rule #1.

• 6. Hydroxides are insoluble except for those in rule #1, and Ba 2+ .

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Conductivity of Solutions Demonstration

Brief description of demonstration

Electrodes of a bulb-type conductivity tester are submerged into acid solutions of increasing strength but
equal concentration. The bulb glows brighter with the increasing acid strength.

Materials

100 mL distilled water

100 mL 0.1 M hydrochloric acid, HCI

100 mL 0.1 M acetic acid, HC 2 H 3 2

100 mL 0.1 M citric acid, H 3 C 6 H 5 7

100 mL 0.1 M malonic acid, H 2 C 3 H 2 04

100 mL 0.1 M ascorbic acid, HC e H 7 O e

100 mL 0.1 propanoic acid, HC3H5O2

100 mL 0.1 glycine, HC 2 H 4 2 N

100 mL alanine, HC 3 H 6 2 N

2 mL universal indicator solution, 1-10 pH range

9 250 mL beakers

disposable pipette

8 stirring rods

conductivity tester, light bulb type

wash bottle

400 mL beaker

Procedure

Label each of the beakers with the appropriate acid names (or distilled water). Pour 100 mL of each acid
into separate beakers. Add 4 drops of universal indicator solution to each beaker and stir. Arrange the

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beakers in order of the spectrum, from red to yellow. A white background behind the beakers will help color
definition. Place the electrodes of the conductivity tester into the glycine solution. Plug the conductivity
tester in. The bulb will glow dimly. (Darken the room if necessary to see it.) Unplug the conductivity
tester, and rinse the electrodes with a wash bottle into a 400 mL beaker. Repeat this procedure with
the other acid solutions, noting that the conductivity and thus the brightness of the bulb increases with
increasing acid strength.

Hazards

All of the acid solutions are corrosive, HCl especially so because it is a strong acid. Avoid contact. The
conductivity tester is a considerable electrical shock hazard, especially with the solutions of electrolytes.
Make sure that it is unplugged before handling it.

Disposal

Rinse each acid solution down the sink with a 100 fold excess of water.

Image Sources

(i) •

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Chapter 17

Acids and Bases

17.1 Arrhenius Acids

Lesson Objectives

• Define an Arrhenius acid and know some examples of acids.

• Define operational and conceptual definition.

• Explain the difference between operational and conceptual definitions.

• Describe the properties of acids.

• Describe some of the reactions that acids undergo.

Introduction

In previous chapters you may remember learning about electrolytes and you may even remember reading
about Svante Arrhenius. Arrhenius worked a great deal with specific types of electrolyte solutions known
as acids and bases. He set the groundwork for our current understanding on acid-base theory. We will
begin our study of acids and bases with Arrhenius's theories starting with his famous definitions. This was
quite an accomplishment for a scientist in the late 19' ft century with very little technology but with the
combination of knowledge and intellect available at the time. Arrhenius led the way to our understanding
of how acids and bases differed, their properties, and their reactions.

We may not realize how much acids and bases affect our lives. Have you ever thought of drinking a can
of soda pop and actually drinking acid? Have you looked at bottles of household cleaners and noticed
what the main ingredients were? Have you ever heard a shampoo commercial and heard them say that the
shampoo was "pH balanced" and wondered what this meant and why it is so important for hair? Thanks
to the beginning work of Arrhenius in the latter part of the 19 century, we started to learn about acids
and bases; our study continued and is constantly growing. Let's begin our study of this wonderful branch
of chemistry.

Definition of Arrhenius Acid

Take a look at all of the following chemical equations. What do you notice about them? What is common
for each of the equations below?

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HCl(aq) -> H+ {a q ) + CI {aq) (Equation 1)

Hydrochloric acid

HN0 3(aq ) -> H + (aq) + N0 3 - {aq) (Equation 2)

Nitric acid

HCl0^ aq) -> H + {aq) + ClO^(a q) (Equation 3)

Perchloric acid

One of the distinguishable features about acids is the fact that acids donate H + ions in solution. If you
notice in all of the above chemical equations, all of the compounds ionized to produce H + ions. This is
the one main, distinguishable characteristic of acids and the basis for the Arrhenius definition of acids.
Arrhenius defined an acid as a substance that produces H + ions in solution.

Most acids can be easily identified because their formula begins with H. Notice this is the case with the
three acids in the equations above. This, of course, is not always the case. Sometimes acids are written
a little differently so that the H is not written first. Let's look at the chemical equation for acetic acid
(Equation 4). It can sometimes be written in a different manner (Equation 5), which is typical for weak
acids.

HC 2 H 3 2 (aq) -> H+ {a q ) + C 2 H 3 2 ~ ( aq ) (Equation 4)

CH 3 COOH (aq) -» H+ (aq) + CH 3 COO- {aq) (Equation 5)

Thus, there are two ways you could identify an acid. You could check to see if the compound formula
begins with an H, this would be a primary indicator. Then write the ionization equation, if the ionization
equation reveals that the H + ion is released, the compound is definitely an acid.

Sample question 1: Which of the following compounds are acids? For those that are acids, write the
ionization reaction.

(a) H 2 SO A

(b) NaOH

Solution:

(a) H 2 S 04 looks like it is an acid because the formula begins with an H. We check by writing the ionization
equation and see that the compound dissociates to give H + ions and therefore is definitely an acid.

H 2 S0 4 ( ag ) -» 2 H ( aq) + S0 4 "(^

(b)NaOH has Na + as a cation, not H + (or starts with a cation other than H + ) and is therefore not an
acid. By writing the dissociation equation we see that NaOH is definitely not an acid.

NaOH {s) -> Na + {aq) + OH~ {aq)

(c) CqH^COOH does not start with hydrogen but when we write the ionization equation, we reveal that
the compound ionizes to give H + ions in solution and is therefore an acid.

C 6 H 5 COOH( aq j t? H+ (aq) + CqH 5 COO~ ( aq )
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Properties of Acids

Acids are a special group of compounds because it has been found that they have their own set of properties.
This helps to identify them from other compounds. Thus, if you had a number of compounds that you
were wondering whether these were acids or otherwise, you could identify them by their properties. But
what exactly are the properties? Think about the last time you tasted lemons. Did they taste sour, sweet,
or bitter? Lemons taste sour. This is another property of acids. Another property of acids is that they
turn blue litmus paper red. Litmus paper is an indicator paper that is used to identify whether a substance
is an acid or a base. If blue litmus paper turns red when it is dipped into a solution, then the solution is
an acid. Figure ?? shows litmus paper and its reaction with an acid solution.

The properties of acids such as sour taste and turning blue litmus red are parts of the operational definition
of acids. Operational definitions describe how the acids behave. Operational definitions differ from
conceptual definitions because with conceptual definitions, there is an attempt to explain why the acid
is behaving the way it is. Let's look at the reaction between hydrochloric acid and sodium hydroxide.
Hydrochloric acid ionizes and produces an H + ion.

HCl

(aq)

i a i)

+ cr

(aq)

(Equation 1)

Hydrochloric acid

Sodium hydroxide, on the other hand, dissociates to produce the sodium ion and the hydroxide ion. Look
at Equation 6.

NaOH^ -> Na + ^ + OH~^ (Equation 6)

When hydrochloric acid and sodium hydroxide react, we get the following chemical equation (Equation 7).

HCh aq \ + NaOH( aq -j — > NaCl( aq j + //20(l) (Equation 7)

Notice how the products of Equation 7 are table salt and water. When an acid (HCl) reacts with a base
(NaOH), a salt and water are produced. Therefore the acid has neutralized the base. The conceptual
definition for the neutralization would state that the H + ions and the OH ions react to form neutral
water molecules. This is true for all reactions between acids and bases.

Sample question: Write the reaction between HNO3 and KOH. Explain how the acid is neutralizing the
base.

Solution:

HN0 3{aq) + KOH {aq) -* KN0 3{aq) + HOH {L)

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As with the case with HCl and NaOH, the H + ions from HN0 3 r aq \ and the OH~ r aq \ ions from the KOHi aq \
combine to form neutral water. This means that the acid has neutralized the base. If we were to write the
total ionic equation for this reaction, we would see:

Total Ionic: H+ {aq) + N0 3 ~ (aq) + K+ {aq) + OH~ {aq) -> K + {aq) + N0 3 " {aq) + H 2 {L)

The net ionic equation then reveals that the acid neutralizes the base. Net ionic: H + t aq \ + OH~ i aq \ — » H 2 On\

Acids React with Metals

We have learned that part of the conceptual definition of acids is that they neutralize bases. Another part
of this definition is that they react with metals to produce hydrogen gas. Look at the chemical reactions
below. What type of reactions are these?

Z« w + 2 HClfa) -> ZnCl 2 ( aq ) + H 2 ( g ) (Equation 8)

Mg {s) + 2 HCl {aq) -> MgCl 2{aq) + H 2{g) (Equation 9)

Ba( s ) + 2 HCl( aq j — > BaClzUq) + #2(g) (Equation 10)

Notice that each of these reactions is a single replacement reaction. Single replacement reactions involve
the reaction between a single element and a compound. In the cases shown above, the single displacement
reactions all involve a metal reacting with an acid. What do you notice that is the same in the product
side of the equation for all three equations? They all produce hydrogen gas (H 2 ). This is another part of
the conceptual definition of acids. Acids react with most metals to produce hydrogen gas.

Sample question: Write the reaction between:

(a) magnesium and sulfuric acid.

(b) calcium and acetic acid
Solution:

(a) Mg (s) + H 2 SO A{aq) -> MgSO A{aq) + H 2(g)

(b) 2 CH 3 COOH {aq) + Ca {s) -» Ca(OOCCH 3 ) 2{aq) + H 2{g)

Acids in Our Environment

Acids are present in our everyday lives. Think about the last time you took an aspirin or a vitamin C
tablet. Aspirin is acetylsalicylic acid while vitamin C is ascorbic acid; both are acids that can produce H +
ions when ionizing in water. Acetic acid {HC 2 H 3 2 ) is a component of vinegar, hydrochloric acid {HCl) is

stomach acid, phosphoric acid (//3PO4) is commonly found in dark soda pop, sulfuric acid (H2SO4) is used
in car batteries and formic acid {HC0 2 H) is what causes the sting in ant bites. The list goes on and on.
We interact with acids on a daily basis so some knowledge of their properties and interactions is essential.

Sample question: Write the names of three common acids other than the ones listed here.

Solution:

Hydrofluoric acid, HF - used to etch glass

Nitric acid, HNO3, - used to etch metals

Citric acid, C 3 H§{COOH) 3 - sour taste in citrus fruits

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Lesson Summary

• Arrhenius defined an acid as a substance that donates H + ions when dissociating in solution. Oper-
ational definitions describe observed properties of how acids behave.

• Conceptual definitions describe why acids behave the way they do.

• Properties of acids include: affect indicators (turn blue litmus paper red), taste sour, neutralize bases,
react with metals to produce hydrogen gases.

Review Questions

1. Explain the difference between a conceptual definition and an operational definition.

2. What are the properties of acids? Give a common example.

3. Which statement best describes a characteristic of acid solutions?

(a) They react with some metals to form hydrogen gas.

(b) They turn red litmus paper blue.

(d) They are made from non-metal oxides.

4. Which of the following is the Arrhenius definition of an acid?

(a) An acid is a substance that donates protons.

(b) An acid is a substance that accepts protons.

(d) An acid is a substance that reacts with water to form H + ions.

5. Which of the following will react with acids and produce hydrogen gas?

(a) chlorine

(b) ammonia

(d) magnesium

6. Write the reaction for each of the following:

(a) hydrofluoric acid + sodium hydroxide

(b) potassium hydroxide + hydrogen sulfide

(d) zinc + hydrochloric acid

Further Reading / Supplemental Links

• http://en.wikipedia.org/wiki

Vocabulary

Arrhenius acid A substance that produces H + ions in solution.

operational definitions Definitions that describe how something behaves, (i.e. the operational defini-
tion of acids includes tastes sour and turns blue litmus red).

conceptual definitions Definitions that describe why something behaves the way it does. (i.e. the
conceptual definition of acids includes reacting with bases to neutralize them) .

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17.2 Strong and Weak Acids

Lesson Objectives

• Distinguish between strong and weak acids.

• Identify strong and weak acids from given choices.

• Describe how strong and weak acids differ in terms of concentrations of electrolytes.

Introduction

A great number of people associate a strong acid with its ability to react with skin, essentially "melting '
it away from bone. It was only recently on a popular crime show that this very acid chemistry know-how
was used as a method for a crime. This crime show used sulfuric acid. Why sulfuric acid and not acetic
acid? What makes the difference? How can we tell if an acid is strong or weak? How does this relates to
the electrolyte lesson we have learned previously, considering that the acids are a combination of the H +
cation and the anion? The answers to these questions and more will be found in the lesson that follows.

The Hydronium Ion

As has been discussed in an earlier chapter, ions in solution are hydrated (Figure 17.1). That is, water
molecules are attached to the ions by the attraction between the charge on the ion and oppositely charged
end of the polar water molecules. A positive ion in solution will be surrounded by water molecules with
the negative ends of the polar water molecule oriented toward the positive ion. A negative ion in solution
will be surrounded by water molecules with the positive ends of the water molecules oriented toward the
negative ion. When we write the formula for these ions in solution, we do not show the attached water
molecules. It is simply recognized by chemists that ions in solution are always hydrated.

Figure 17.1: Hydrated ions in solution.

The hydrogen ion in solution is also hydrated in the same as all other ions. Many chemists feel that it
is important in the case of hydrogen ion to show that it is attached to water. They choose not to show
that the hydrogen ion is surrounded by four water molecules, but show only that it has one water molecule
attached. When you add a hydrogen ion, H + to one water molecule, H2O, the result is a positively charged
ion consisting of three hydrogens and one oxygen, H^O + . This ion has been given the name hydronium
ion. Many authors show this hydronium ion in every case where a hydrogen ion is represented in an
equation, but other authors show it only sometimes. Expressing the hydrogen ion as a hydronium ion

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complicates equations somewhat because if the hydrogen ion is shown as a hydronium ion, then the other
side of the equation must contain a water molecule to balance the equation.

HCl

(aq)

+ cr

(aq)

(not showing hydronium)

HCl(aq) + #20(L) -> H 3 ( ag ) + CI ( aq )

(showing hydronium)

You will need to recognize that sometimes the hydrogen ion is shown simply as a hydrogen ion and
sometimes, it is shown as a hydronium ion and you will need to be able to deal with it in either form.

Strong Acids

In an earlier chapter, ionization was defined as certain covalently bonded molecules reacting with water and
producing ions in solution. Ionization reactions involve chemical changes. It is the amount of ionization
that is essential in determining if an acid is strong or weak. Strong acids are defined are ones that
completely ionize or undergo 100% ionization (See Equation 1).

HCl (g) + H "zO(L) -* H 3 (aq) + CI ( aq )

(Equation 1)

Notice in equation 1 that there is a single arrow separating the products from reactants. This single arrow
indicates that when the reaction has stopped, there are no HCh g \ molecules remaining in the solution, only
H^ (aq) i° ns an d Cl~f aq \ ions. This is characteristic of a strong acid.

There are only six common strong acids. These acids are shown in Table 17.1. Each of the acids found in
this table, like HCl, completely ionize in water.

Table 17.1: Strong Acids

Name

Symbol

Hydrochloric Acid
Hydrobromic Acid
Hydroiodic Acid
Nitric Acid
Perchloric Acid
Sulfuric Acid

HCl

HBr

HN0 3

HClOi

H 2 SOi

Weak Acids

Unlike their strong acid counterparts, weak acids do not ionize 100%. The less ionization that takes place,
the weaker the acid since there will be fewer H + ions in solution. For example, acetic acid ionizes only
about 5% meaning that when acetic acid is placed in water, only about 5% of the acetic acid molecules
separate into H + ions and C2//3O2 ions (See Equation 2).

HC2H 3 2 ( aq ) ^=> H (aq) + C2H3O2 (aq)

(Equation 2)

Notice, as well, that in the above equation our arrow has been replaced with a double arrow indicating
that the reaction reaches equilibrium. When this reaction reaches equilibrium, the container holds mostly
acetic acid molecules and a little bit of hydronium (or H + ) ions and a little bit of acetate ions.

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Let's look at citric acid. Citric acid, C^figO-?, is commonly found in everyday products like lemons and
limes, and even soft drinks. It is the substance responsible for making the sour taste of these foods and
drinks. Citric acid only ionizes a little more than 3% and is therefore classified as a weak acid. If we were
to write an ionization reaction for citric acid, it would appear as written in Equation 3.

C 6 Hg0 7 (aq) + H 2 O tyVj i? H 3 + (a? ) + CqH 7 7 ~ \ aq ) (Equation 3)

Notice how, as with Equation 2, the double arrow indicates the reaction has reaches equilibrium. As with
all weak acids, when this reaction reaches equilibrium, there is mostly C&HgO-jt aq \ and some hydronium
ion and some citrate ion, C%HjOf~ ' r aq \ in the solution.

Remember that all acids that are not one of the six listed in Table 1 are weak. These weak acids do not
completely ionize in water. Even though these weak acids are very soluble, they dissolve as molecules and
only a few of the molecules break into ions in the solution.

Sample question: Write ionization equations for only those acids that are weak.

(a) Sulfuric acid (H2SO4)

(b) Hydrofluoric acid (HF)

(a) H2SO4 is a strong acid (one of the six).

(b) HF is a weak acid (not one of the six): HF{aq) t> H + (aq) + F~{aq)

Strong and Weak Electrolytes

Non-electrolytes have been described as solutions that do not conduct electricity and electrolytes are those
that do conduct electricity. However, electrolytes do have varying degrees of strength. If a solution has a
large number of ions present in it, it is called a strong electrolyte whereas an electrolyte solution that has
only a few ions present is called a weak electrolyte. A strong acid completely ionizes in water solution,
there are lots of ions in solution and the ions make the solution a strong electrolyte. A weak acid produces
only a few ions in solution and therefore is classified as a weak electrolyte. Such a solution will only conduct
a small electric current.

Lesson Summary

• Strong acids undergo 100% ionization in water (i.e. hydrochloric acid, HCL). Weak acids undergo
less than 100% ionization (i.e. acetic acid, //C2//3O2).

• All acids that are not one of the six are weak and also if they are not one of the six, do not completely
ionize in water. *Weak electrolytes are solutions that conduct electricity to a lesser extent than strong
electrolyte solutions.

Review Questions

1. What is the difference between a strong and weak acid? Show an example of each.

2. In terms of electrolyte solutions, how would you distinguish between a strong acid and a weak acid?

3. All of the following are weak acids except?

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(a) HCIO3

(b) HC 2 H 3 2

(d) //C/

4. Which compound is a strong acid?

(a) HCl0 2 {aq)

(b) H 2 C0 3 (aq)

(d) perchloric acid

5. Which one of the following compounds is not a strong electrolyte?

(a) CH 3 COOH(aq)

(b) HCl0 4 {aq)

(d) NaOH(aq)

6. Which of the following is usually referred to as strong acid in water solution? Write the ionization
reactions.

(a) HF

(b) HN0 2

(d) HS0 4 "

(e) //7V0 3

(f) //C/O4

Further Reading / Supplemental Links

• http://en.wikipedia.org/wiki

Vocabulary

strong acid Acids that completely ionize or undergo 100% ionization in solution (i.e. HCL).

weak acids Acids that do not completely ionize or undergo 100% ionization in solution (i.e. HC 2 H 3 2 ).

17.3 Arrhenius Bases

Lesson Objectives

• Define an Arrhenius base and know some examples of bases.

• State the properties of bases.

• Describe the neutralization reaction that bases undergo.

Introduction

Arrhenius broke ground in our understanding of acids and bases. He was the first to provide us with a
definition from which we could identify an acid from a base. We have learned earlier the definition of an

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acid and now we will extend this definition to include bases. Keep in mind that Arrhenius came up with
these theories in the late 1800s so his definitions came with some limitations. These limitations will be
expanded upon later on in the unit. For now we will focus on his definitions.

Bases Release OH-in Solution

Arrhenius defined an acid as a substance that releases H + ions in solution. In contrast, he defined a base
as a substance that releases OH ions in solution. Bases are ionic substances made up of a cation and an
anion, of which the anion is the OH~ ion. It should be noted that very few of the hydroxides are actually
soluble. If you recall from a previous chapter, only the alkali metals and Ba 2+ ions are soluble. Therefore,
few of the basic solids will result in solutions when dissolved in water. One of the metal hydroxides that
is very soluble is NaOH. The dissolving equation for NaOH is shown in Equation 1.

NaOH^ -* Na + ( aq ) + OH~u^\ (Equation 1)

Barium hydroxide produces a similar reaction when dissociating in water (Equation 2).
Ba(OH) 2{s) -» Ba 2+ {aq) + 2 OH~ {aq) (Equation 2)

The production of OH" ions is part of the conceptual definition of bases according to the Arrhenius
definition of bases.

Sample question: Write the chemical equation for the reaction of the following bases in water.

(a) Lithium hydroxide (LiOH)

(b) Sodium hydroxide (NaOH)

(a) LiOH (s) -> Li + {aq) + OH {aq)

(b) NaOH {s) -> Na + {aq) + OH~ {aq)

Properties of Bases

There is one common base that some may have had the opportunity to taste: milk of magnesia is a slightly
soluble solution of magnesium hydroxide. This substance is used for acid indigestion. Flavorings have been
added to improve the taste, otherwise it would have a bitter taste when you drink it. Other common bases
include substances like Windex, Drano, oven cleaner, soaps and many cleaning other products. Please
note: do not taste any of these substances. A bitter taste is one property you will have to take for granted.

If you notice, one of the common bases is soap. This is an interesting example because another property
of bases is that they are slippery to the touch. When you are washing dishes, you have probably noticed
that the soapy water gets quite slippery. In fact, there may have been times when you dropped a dish or a
glass or a cup because it was too slippery to hold on to. Now you know it isn't your fault - it's the soap's
fault!

As with acids, bases have properties that allow us to identify them from other substances. We have learned
that acids turn blue litmus paper red. It stands to reason then that bases would turn red litmus paper
blue (Figure ??). Notice that the effect of the indicator is the opposite of that of acids.

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As mentioned previously, bases have the ability to neutralize acids. Take for example the reaction between
magnesium hydroxide (milk of magnesia) and HC1 (stomach acid) shown in Equation 3.

Mg(OH) 2{s) + 2 HCl {aq) -» MgCl 2(aq) + 2 H 2 (L)

(Equation 3)

You see in Equation 3 that here again we have an acid and a base producing a salt and water. This is
a similar reaction to the reaction between sodium hydroxide and hydrochloric acid. We will learn more
about these reactions in the next section as well.

Sample question: Write the balanced neutralization reaction between the following acids and bases.

(a) HC10 4 + LiOH

(b) HNO3 + Ba{OH) 2

(a) HClO^aq) + LiOH {aq) -> LiCl0^ aq) + H 2 0( L )

(b) 2 HN0 3{aq) + Ba{OH) 2{aq) -> Ba(N0 3 ) 2{aq) + 2 H 2 {L)

Common Bases

We mentioned above that bases are common to our everyday lives. If you do any of the cleaning in
your home, you would use bases quite frequently. Drano©, used to unclog drains, is a solution of sodium
hydroxide; sodium hydroxide, NaOH, is also used to make some soap; soft soap is often prepared with
potassium hydroxide, KOH. We have already mentioned that magnesium hydroxide, Mg{OH) 2 , is used to
make milk of magnesia. Windex is a water solution containing ammonium hydroxide, NH4OH.

Soaps, as we know, have the ability to dissolve in water but also dissolve oil substances as well. The reason
soap is able to do this is because of its structure. Soaps have long chains of carbon atoms, which make
that part of the molecule nonpolar. This allows it to dissolve other nonpolar substances such as oils. And
then on one end of the molecule is a sodium ion which makes that part of the molecule ionic. This ionic
end of the molecule will be attracted to polar water molecules. This nonpolar-ionic molecule can attach
any nonpolar particles to water molecules, and wash them away.

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CH 3 (CH 2 ) 16

Hydrocarbon end Ionic end

(Water repelling) (Water attracting)

The Structure of Soap.

Lesson Summary

• An Arrhenius base is a substance that releases OH~ ions in solution. Bases turn red litmus paper
blue, have a bitter taste, are slippery to the touch, and can neutralize acids.

• A neutralization reaction between an acid and a base will produce a salt and water. Example:
Mg(OH) 2{aq) + 2 HCl {aq) -* MgCl 2{aq) + 2 H 2 (L) .

• Common bases include soap (NaOH) and Drano (NaOH), soft soap (KOH), milk of magnesia
(Mg(OH) 2 ) and Windex (NH 4 OH).

Review Questions

1 . What is the role of litmus paper for acids and base chemistry?

2. What are the properties of bases? Give a common example.

3. Which statement best describes a characteristic of a base solutions?

(a) They taste bitter.

(b) They turn red litmus paper red.

(d) They are weak electrolytes.

4. Which of the following is the Arrhenius definition of a base?

(a) A base is a substance that donates protons.

(b) A base is a substance that accepts protons.

(d) A base is a substance that reacts with water to form H + ions.

5. Which of the following bases would be a weak electrolyte?

(a) NaOH

(b) Ba(OH) 2

(d) Al(OH) 3

6. Write the balanced neutralization reaction between the following acids and bases.

(a) potassium hydroxide + hypochlorous acid

(b) hydrobromic acid + calcium hydroxide

(d) potassium hydroxide + sulfuric acid

7. Write the net ionic equation for each of the neutralizations reactions in #4.

Further Reading / Supplemental Links

• http://en.wikipedia.org/wiki
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Vocabulary

Arrhenius base A substance that produces OH" ions in a solution.

17.4 Salts

Lesson Objectives

• Describe the formation of salts in neutralization reactions in terms of Arrhenius theory.

• Identify acidic, basic, and neutral salts from neutralization reaction.

Introduction

Neutralization is a reaction between an acid and a base which produces water and a salt. The general
reaction for the neutralization reaction is shown below.

acid + base — » salt + water

In this lesson, after reviewing the concept of ionic compounds, we will examine neutralization reactions
and then look at the different type of salts that can be formed from acids and bases as they react in
neutralization reactions.

Ionic Compounds

Ionic compounds are those formed between metal cations and nonmetal anions. When you take a look at
the periodic table, the compounds with the most ionic character are those that are at opposite sides of the
table. For example, sodium chloride will have more ionic character than zinc chloride. Look at Figure
17.2 to see the positions of sodium, Na, zinc, Zn, and chlorine, CI.

::•■

;..

Figure 17.2: The positions of sodium, zinc, and chlorine in the periodic table.

When ionic compounds form between metal cations and nonmetal anions they transfer electrons and form
charged particles and are capable of forming electrolyte solutions. If you notice, this is all similar to the
descriptions of acids and bases. Acids are a combination of a hydrogen cation and a nonmetal anion.
Examples include HCl, HNO3, and //C2//3O2. Bases can be a combination of metal cations and nonmetal
anions. Examples include NaOH, KOH, and Mg{OH)2-

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Salts

Salts that are ionic compounds result when an acid and a base react in a neutralization reaction.

Acid + Base -» Salt + Water

It is this salt that is formed that is the ionic compound. Let us examine some examples of the reactions
we have looked at to see some of the salts and their ionic character. Take a look at Equations 1,2, and 3.
Each of these equations has an acid and a base producing a salt and water.

HCl {aq) + LiOH {aq) -> LiC\ aq) + H 2 {L) (Equation 1)

2 HBr {aq) + Ba(OH) 2{aq) -> BaBr 2{aq) + 2 H 2 (L) (Equation 2)

2 HF (aq) + Mg(OH) 2(aq) -» MgF 2{aq) + 2 H 2 {L) (Equation 3)

Notice that in Equation 1, lithium chloride (LiCl) shows Li + ions and the Cl~ ions are the furthest away
from each other on the periodic table so, generalizing, they will produce the salt with the most ionic
character. Magnesium fluoride in Equation 3 has Mg 2+ ions and F— ion, which are closer to each other on
the Periodic Table and therefore will have a little bit less ionic character.

All acid-base reactions produce salts. According to the conceptual definition of the Arrhenius acid and
base, the acid will contribute the H + ion that will react to neutralize the OH~ ion, contributed by the
base, to produce neutral water molecules. The anion from the acid will combine with the cation from the
base to form the ionic salt. Look at Equations 4 and 5 below to see the conceptual definition in action.

HClO^aq) + NaOH( aq) -> NaCl0 4 ^ aq) + HOH^ L) (Equation 4)

H 2 S <aq) + 2KOH {aq) -> K 2 S 4{aq) + 2 HOH (L) (Equation 5)

(Note: HOH = H 2 0)

No matter what the acid or the base may be, the products of this type of reaction will always be a salt
and water. Aside from the fact that Arrhenius said that the H + ion will neutralize the OH~ ion to form
water, we also know that these reactions are double displacement reactions and will therefore have their
cations exchanging anions.

To write this with a total ionic equation:

H+ {aq) + Cl04~(aq) + Na+ \aq) + OH ( my ) -» Na + \ aq} + CIO4 \ aq) + H 2 0( L )
Or, since Na + and ClO^~ are spectator ions, the net ionic equation is:

H+ {aq) + OH ' ( aq) -> H 2 0(l)

This is the net ionic equation for all neutralization reactions.
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The Hydrolysis of Salts
The Hydrolysis Reaction

When a salt is dissolved in water, it is possible for the solution to be neutral, acidic, or basic. If a solution
is to be acidic, it must contain more hydrogen ions than hydroxide ions and for the solution to be basic, it
must contain more hydroxide ions than hydrogen ions. The reason that a dissolving salt may produce non-
neutral solutions is that the cation or anion from the salt may not dissociate 100% when in the presence
of hydrogen or hydroxide ions. Water, itself, dissociates very slightly into hydrogen and hydroxide ions.
Therefore, when a salt is dissolved in water, there are some hydrogen and hydroxide ions available in the
solution. Consider, first, the solution produced when the salt KBr dissolves in water. There will be four
ions present in the solution.

K + + Br + H + + OH ±5?

When potassium ions in the solution come into contact with hydroxide ions, if the ions were to join together,
the molecule formed would be KOH, which is a strong base, and would, therefore, immediately dissociate
back into ions. Similarly, if the bromide ions come into contact with hydrogen ions, the molecule formed
would be HBr, which is a strong acid, and would immediately dissociate back into ions. Having sodium
and chloride ions in a water solution would not cause a reaction.

Consider, now, the solution produced when the salt NH4CI is dissolved in water. There will be four ions
present in the solution.

NH+ + CI + H + + OH ±5?

When hydrogen ions come into contact with chloride ions, if they join together, the resultant molecule
would be HCl, which is a strong acid, and therefore the HCl would immediately dissociate back into
the ions. When NH^ + ions come into contact with OH ions, however, the resultant molecule would be
NH4OH, which is a weak base and therefore, does not dissociate very much. Therefore, when ammonium
chloride is dissolved in water, a reaction occurs.

NH 4 + + Cr + H + + OH ±* NH A OH {aq) + Cl~ + H +

The ammonium hydroxide dissociates very little, so we would have mostly un-dissociated ammonium
hydroxide molecules in solution with hydrogen and chloride ions. The hydrogen ions in this final solution
would cause the solution to be acidic. Thus, dissolving ammonium chloride in water produces an acidic
solution.

By a similar process, dissolving sodium acetate, NaC 2 H 3 2 , in water will produce a basic solution. When
the sodium acetate is dissolved in water, four ions will be present in the solution.

Na + + C 2 H 3 2 +H+ + OH £??

If sodium ions contact hydroxide ions, the substance formed would be a strong base which would immedi-
ately dissociate. If hydrogen ions contact acetate ions, however, the molecule formed would be acetic acid,
which is a weak acid and the ions would NOT dissociate. Therefore, when sodium acetate is dissolved in
water, a reaction will occur as shown below.

Na + + C 2 H 3 2 + H + + OH £5 HC 2 H 3 2{aq) + Na + + OH

The resultant solution has excess hydroxide ions and therefore, is basic. The dissolving of sodium acetate
salt in water produces a basic solution.

The reactions between some of the ions in the salts and water are called hydrolysis reactions.

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Neutral, Acidic, and Basic Salts

We have talked a little already about the fact that there are not just strong acids but there are also weak
acids as well. The difference between them is the percent ionization. The same is true for bases. Strong
bases dissociate 100% and weak bases do not. Table 17.2 shows all of the strong acids and bases, all of
the rest of the acids and bases are weak.

Table 17.2: Strong Acids and Bases

Strong Acid

Formula

Strong Base

Formula

Lithium hydroxide

LiOH

Sodium hydroxide

NaOH

Potassium hydroxide

KOH

Rubidium hydroxide

RbOH

Cesium hydroxide

CsOH

Calcium hydroxide

Ca(OH) 2

Strontium hydroxide

Sr{OH) 2

Barium hydroxide

Ba(OH) 2

Hydrochloric Acid

HCl

Hydrobromic Acid

HBr

Hydroiodic Acid

Nitric Acid

HN0 3

Perchloric Acid

HClOi

Sulfuric Acid

H 2 S0 4

So why is this important to us here? The information in the table helps us to determine what type of salt
is formed in an acid-base reaction. For example if we have a reaction between a strong acid and a strong
base we form a neutral salt. It is like a power struggle between the acid and the base. Since both are
strong, there is no compound with more power and thus the salt ends up being neutral. If, however, we
have a reaction between a weak acid and a strong base (Equation 6), the result would be a basic salt.

HC 2 H 3 2 (aq)

NaOH (aq)

NaC 2 H 3 2 (aq)

#20(X)

(Equation 6)

Acetic acid

sodium hydroxide

sodium acetate

water

(weak acid)

(strong base)

(basic salt)

When the basic salt is dissolved in water, a reaction called hydrolysis takes places in which extra hydroxide
ions, OH 1 —, are produced from the salt and the water molecules. Since the hydroxide ions come from a
salt, it is called a basic salt.

A similar situation will occur when we have a strong acid reacting with a weak base. When a strong acid
reacts with a weak base, an acidic salt is formed.

HCl( aq )

Hydrochloric acid
(strong acid)

+ NH 4 OH {aq)

ammoniun hydroxide
(weak base)

NHiCl^q) + H 2 {L) (Equation 7)

ammonium chloride water

(acidic salt)

Again, a hydrolysis reaction takes place. The salt will react with water molecules and produce excess
hydrogen ions, H + , in solution and is therefore referred to as an acidic salt.

Sample question: Complete the following neutralization reactions and identify the type of salt produced.

(a) H 2 S0 4(ag) + Ba(OH) 2{aq) -»?

(b) HCOOH {aq) + Ca(OH) 2 (aq) -»?
Solution:

(a) H 2 S0 4{aq) + Ba{OH) 2{aq) -» BaSO i{aq) + 2H 2 {L)

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Strong acid + strong base = neutral salt

(b) 2 HCOOH (aq) + Ca{OH) 2(aq) -» Ca(HCOO) 2{aq) + 2 Zf 2 (L)

Weak acid + strong base = basic salt

It can also be determined which acid and base was used to form the salt and from this you could determine
if your starting reaction was a strong acid or a weak acid, a strong base or a weak base. Watch how this
works. You were given that the product salt was calcium nitrate, Ca(NOs)2- Remember that there is a
double displacement reaction that forms the salt so we can write parts of the reaction:

H? + ?0H *■ Ca(N0 3 ) 2 + HOH

Calcium must come from the Nitrate must come from the

base. Therefore, the base acid. Therefore, the acid

must be Ca(0H} 2 . mustbeHNQ s .

Ca(0H) 2 is a strong base. HN0 3 is a strong acid.

Therefore, the neutralization reaction would have been:

2 HN0 3 + Ca(OH) 2 -» Ca(N0 3 ) 2 + 2 HOH (Equation 8)

So this salt would have been produced from a strong acid-strong base reaction!

Let's try another. Consider the salt copper (II) chloride (CuCl 2 ).

The copper would have come from the base, Cu(OH) 2 , which is a weak base. The chloride would have
come from the acid, HCl, which is a strong acid.

2 HCl + Cu(OH) 2 -> CuCl 2 + 2 HOH (Equation 9)

Therefore the reaction is a strong acid weak base reaction!

Lesson Summary

• Ionic compounds are those formed between metal cations and nonmetal anions. Acids are a com-
bination of hydrogen and a nonmetal anion. Bases can be a combination of metal cations and
nonmetal anions. Ionic compounds result when an acid and a base react in a neutralization reaction
(Acid+ Base -> Salt + Water).

• According to the conceptual definition of the Arrhenius acid and base, the acid will produce the H +
ion, which will react to neutralize the OH' ion produced by the base to produce the neutral water.
The other product will be the ionic salt. A strong acid + a strong base in an acid/base neutralization
reaction will form a neutral salt.

• A strong acid + a weak base in an acid/base neutralization reaction will form an acidic salt. A weak
acid + a strong base in an acid/base neutralization reaction will form a basic salt.

Review Questions

1. How do an acid and a base fit the definition of an ionic compound? Use examples in your answer.

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2. Explain neutralization reactions in terms of Arrhenius theory. Use an example in your answer.

3. Which salt will form a basic solution when dissolved in water?

(a) KN0 3

(b) CaCl 2

(d) NaN0 2

4. Which salt will form an acidic solution when dissolved in water?

(a) copper(II) sulfate

(b) sodium acetate

(d) sodium cyanide

5. Milk of magnesia is a common over- the - counter antacid that has, as its main ingredient, magnesium
hydroxide. It is used by the public to relieve acid indigestion. Acid indigestion is caused by excess
stomach acid being present. Since a stomach upset is caused by excess hydrochloric acid, this tends
to be a quite painful affliction for people. Write the balanced chemical equation for the reaction
between milk of magnesia and hydrochloric acid. What type of reaction is this? What type of salt is
formed?

6. Complete the following neutralization reactions and identify the type of salt produced.

(a) H2S Oi(aq) and NaOH{aq) — >

(b) HN0 3 (aq) and NH A OH{aq) ->

(d) CH 3 COOH(aq) and KOH{aq) ->

(e) HCl(aq) and KOH(aq) ->

Further Reading / Supplemental Links

• http : //en . wikipedia . org/wiki/Salts

Vocabulary

basic salt A salt formed in a neutralization reaction between a weak acid and a strong base.

acidic salt A salt formed in a neutralization reaction between a strong acid and a weak base.

neutral salt A salt formed in a neutralization reaction between a strong acid and a strong base or a
weak acid and a weak base.

Labs and Demonstrations for Acids and Bases
Teacher's Pages for Hydrolysis of Salts

Investigation and Experimentation Objectives

In this activity, the student will predict the acidity or basicity of hydrated salts and then carry out
laboratory activities to compare the relationships between evidence and explanations.

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Lab Notes

Preparation of Solutions:

200 mL of each solution should be more than enough to complete this lab. To prepare each solution, mass
the specified amount of reagent, dissolve it in 150 mL of water, and dilute the resulting solution to 200 mL:

CuSOi- 5H2O : 5.0 grams

Ca(NOs)2 : 3.3 grams

K3PO4 : 4.2 grams

KCl : 1.5 grams

NaBr : 2.1 grams

Na2S : 1.6 grams

(NH 4 ) 2 C0 3 : 1.9 grams

Na<iCrOt± : 3.2 grams

MgBr 2 : 3.7 grams

NaCl : 1.2 grams

Use care when opening the container of (NH 4 )2C0 3 . It undergoes decomposition over time, and outgases
NH 3 , which collects in the container. Upon opening the NH 3 diffuses out rapidly, and is very irritating to
eyes and skin. Live and learn.

Answers to Pre-Lab Questions

1. Write dissociation equations for the following salts:

(a
(b

(c
(d

(0
(f
(g
(li
(i
(j

CuS0 4 ^Cu 2+ + SO\ +
Ca{N0 3 ) 2 -» Ca 2+ + 2 N0 3
K 3 P0 4 -> 3 K+ + POf-
KCl -> K + + Ct
KBr -» K + + Br
Na 2 S -^2 Na + + S 2 -
(NH 4 ) 2 C0 3 -> 2 NH+ + COf-
Na 2 CrOi -^ 2 Na + + CrOf
MgBr 2 -» Mg 2+ + 2 Br
NaCl -^ Na + + CI

2. NaC e H 5 C0 2 + H 2 -^ HC 6 H 5 C0 2 + Na + + OH solution will be basic

Lab — Hydrolysis of Salts

Background Information

A salt is an ionic compound containing positive ions other than hydrogen and negative ions other than
hydroxide. Most salts will dissociate to some degree when placed in water. In many cases, ions from the
salt will react with water molecules to produce hydrogen ions, H + , or hydroxide ions, OH . Any chemical
reaction in which water is one of the reactants is called a hydrolysis reaction. Salts are usually formed from
the neutralization reaction between an acid and a base. A salt formed from a strong acid and a strong base
will not undergo hydrolysis. The resulting solution is neutral. An example of such a salt is KBr, formed
from a strong acid, HBr, and a strong base, KOH.

Salts formed from the reaction of a strong acid and a weak base hydrolyzes to form a solution that is

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slightly acidic. In this kind of hydrolysis, the water molecules actually react with the cation from the weak
base. For example, when ammonium chloride, NH4CI, hydrolyzes, water molecules react with the NHf
ion:

NH+ + H 2 -> NH4OH + H +

The formation of the H + ion from this reaction makes the solution acidic.

Salts formed from the reaction of a weak acid and a strong base hydrolyze to form a solution that is slightly
basic. In this kind of hydrolysis, it is the anion from the weak acid that actually reacts with the water.
For example, when sodium acetate, NaC 2 H 3 2 , hydrolyzes, water molecules react with the acetate ion:

C 2 H 3 2 +H 2 -> HC 2 H 3 2 + OH

The formation of the OH ion from this reaction makes the solution basic. Salts formed from a weak acid
and weak base produce solutions that may be slightly acidic, slightly basic, or neutral, depending on how
strongly the ions of the salt are hydrolyzed.

In this experiment you will test several different salt solutions with pH paper and phenolphthalein solution
to determine their acidity or basicity.

Purpose

To determine the relative acidity or basicity of various salt solutions, and thus predict whether hydrolysis
occurred, and if so, what the reaction products are.

Pre-Lab Questions

1. Write dissociation equations for the following salts:

(a
(b

(c
(d

Copper(II) sulfate
Calcium nitrate
Potassium phosphate
Potassium chloride
Potassium bromide
Sodium sulfide
Ammonium carbonate
Sodium chromate
Magnesium bromide
Sodium chloride

(h)
(i)
0)

2. Sodium benzoate is the salt formed in the neutralization of benzoic acid with sodium hydroxide.
Benzoic acid is a weak acid. Write the hydrolysis reaction for the dissolution of solid sodium benzoate,
NaC^H^C0 2 , in water. Will sodium benzoate solution be acidic, basic, or neutral?

Apparatus and Materials

• 10 small or medium sized test tubes, or a micro reaction plate

• Test tube rack

• 0.1 M solutions of cupric sulfate, calcium nitrate, potassium phosphate, potassium chloride, sodium
bromide, sodium sulfide, ammonium carbonate, sodium chromate, magnesium bromide and sodium
chloride

• Universal pH indicator paper, range 0-14

• Phenolphthalein indicator solution

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• 10 mL graduate

• Stirring rod

Safety Issues

The solutions used may be slightly acidic or basic, and as a result can be corrosive or caustic. Use proper
laboratory safety equipment and techniques.

Procedure

1. Obtain a clean, dry micro reaction plate, or 10 test tubes

2. To test tubes 1 through 10 or the reaction plate, add eight to ten drops of the following solutions:

Tube or Well 1: Cupric sulfate
Tube or Well 2: Calcium nitrate
Tube or Well 3: Potassium phosphate
Tube or Well 4: Potassium chloride
Tube or Well 5: Sodium bromide
Tube or Well 6: Sodium sulfide
Tube or Well 7: Ammonium carbonate
Tube or Well 8: Sodium chromate
Tube or Well 9: Magnesium bromide
Tube or Well 10: Sodium chloride

3. Add two drops of phenolphthalein solution to each of the occupied wells of the microplate or test tube.
Record your observations in the data table.

4. Test each solution with pH paper and record your results.
Data

Table 17.3: Data Table
Well # Salt

1
2
3
4
5
6
7
8
9
10

Post-Lab Questions

1. Where a hydrolysis is likely to occur in each of the following, write a net ionic hydrolysis equation. If
no hydrolysis is likely, write NR.

Effect on

Original

Strong

Original

Strong

Indicator

Acid

or Weak
Acid

Base

or Weak
Base

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Cu 2+ + SOf + 2 Zf 2 0-»

Ca 2+ + 2 N<9 3 + 2 H 2 ->

3 K + + POf + 3 H 2 ->

k + + cr + H 2 ->

Ma + + Br + # 2 <9 ->

2 Na + + S 2 ~ + H 2 ->

2 JV/f+ + CO 2 ^ + Zf 2 -»

2 Na + + CrOf" + 2 Zf 2 -»

Mg 2+ + 2 5r" + 2 H 2 -»

Na + + C7 - + Z/ 2 -»

2. How do your observations and /?// readings compare with the expected results based on the equations
for the hydrolysis reactions?

3. What is a spectator ion? Name the spectator ions present in each hydrolysis reaction in this experiment.

4. A salt formed from a strong acid and a strong base produces a neutral solution. A salt of a weak acid
and a weak base may or may not produce a neutral solution. Explain why.

5. Bases make effective cleaning agents, because they can convert grease and oils to a water soluble
substance. Trisodium phosphate (TSP) is a common commercially available cleaner. Give the reaction
TSP undergoes to create a basic solution.

17.5 pH

Lesson Objectives

• Calculate [H + ] for strong acids and [OH~] for strong bases.

• Define autoionization and use it to find [H + ] from [OH~] or to find [OH~] from [H + ].

• Describe the pH scale.

• Define pH.

• Calculate pH from [H + ] or vice versa.

Introduction

A few very concentrated acid and base solutions are used in industrial chemistry and occasionally in
inorganic laboratory situations. For the most part, however, acid and base solutions that occur in nature,
those used in cleaning, and those used in organic or biochemistry applications are relatively dilute. Most
of the acids and bases dealt with in laboratory situations have hydrogen ion concentrations between 1.0 M
and 1.0 x 10~ 14 M.

Expressing hydrogen ion concentrations in exponential numbers becomes tedious and is difficult for those
not trained in chemistry. A Danish chemist named Soren Sorenson developed a shorter method for express-
ing acid strength or hydrogen ion concentration with a non-exponential number. He named his method
pH and while the exact definition of pH has changed over the years, the name has remained.

pH, today, is defined as the negative logarithm of the hydrogen ion concentration.
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pH = -log [H + ]

If the hydrogen ion concentration is between 1.0 M and 1.0 x 10 -14 M, the value of the pH will be between
and 14.

Concentrations of Ions

We have said that a strong acid completely ionizes in solution and therefore in solution there are no intact
acid molecules remaining but only the H + cations and the anions from the acid. Look at Equation 1 as an
example.

HC\ aq) -> H + {aq) + Cr {aq) (Equation 1)

100% ionized

We can extend this concept to give actual concentrations of the hydronium ion for strong acid solutions.
If we had a 0.10 mol/L solution of hydrochloric acid and it completely ionizes, then the concentration of
H + ions that are produced are also 0.10 mol/L. Look at the balancing coefficients for Equation 1. The
balancing coefficients are all ones (1). Therefore the concentrations will all be the following:

HCl {aq) -> H+ {aq) + Cl ~(aq) (Equation 2)

0.10 mol/L 0.10 mol/L 0.10 mol/L

For strong bases, the same calculation can be performed. Since strong bases are 100% dissociated, when
we are given the concentration of the strong base we can then conclude the concentration of the hydroxide
ion. Let's look at the following example. What would be the [OH~] for a solution of Ba(OH)2, knowing
the concentration of Ba{OH)2 is 0.24 mol/L?

Ba(OH) 2{aq) -> Ba 2+ {aq) + 2 OH~ {aq) (Equation 3)

100% dissociated

Notice both the 100% dissociation and the balancing coefficients. Now let's fill in the coefficients. Take
a look at Equation 4 to see how the balancing coefficients have helped determine the [OH~] in the same
manner as they did for finding the [H + ].

Ba{OH) 2{aq) -> Ba 2+ {aq) + 20H~ {aq) (Equation 4)

0.24 mol/L 0.24 mol/L 0.48 mol/L

Sample question: What would the [H + ] be for the ionization of a 0.35 mol/L solution of nitric acid?
Solution:

HN0 3{aq) -> H + {aq) + N0 3 - {aq)

0.35 mol/L 0.35 mol/L 0.35 mol/L

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Relationship Between [H+] and [OH-]

Concentrated hydrochloric acid is found in most high school chemistry labs. This acid is 12.0 mol/L or
12 mol of HCl dissolved in enough water to make 1 L. It is the presence of the water molecules that
allows that hydrogen chloride to ionize and produce the hydrogen ions that end up in solution. Water
also undergoes ionization. Actually, water undergoes autoionization. Autoionization is when the same
reactant acts as both the acid and the base. Look at the reaction below.

#2#(L) + #2<9(L) ^> H 3 + ( aq } + OH'fa)

From experimentation, chemists have determined that in pure water [H + ] = 1 x 10~ 7 mol/L and [0// _1 ] =
1 X 1CT 7 mol/L. In other words, //20(n is autoionizing and so the [H + ] = [OH"] = 1 x 1CT 7 mol/L.

The ionization of water is frequently written as H20n\ t> H + + OH

The equilibrium constant expression for this dissociation would be:

K w = [H+][OH-}

Like other equilibrium constants that are for special reactions, this K is given its own subscript. Since this
equilibrium constant is for the dissociation of water, it is designated as K w . The ion product constant for
water, K w is the product of the hydronium ion and the hydroxide ion concentrations in the autoionization
of water.

We can the calculate K w because we know the value of [H + ] and [OH"] for pure water at 25° C.

K w = [H+][OH-]

K w = (lxl(T 7 )(lxl(r 7 )

K w = 1 x 1(T 14

A further definition of acids and bases can now be made:
when

[//3<9 + ] = [OH~l] (as in water), the solution is neutral
[H%0 + ] > [OH" I] the solution is an acid
[//3<9 + ] < [0// _ l] the solution is a base

To continue with these ideas:

an acid has a [H%0 + ] that is greater than 1 x 1CT 7 and the [OH" 1 ] is less than 1 x 1CT 7

a base has a [O// 1- ] that is greater than 1 x 1CT 7 and the [//30 + ] is less than 1 x 10~ 7

The equilibrium between the water molecules and the H + and OH" ions will exist in all water solutions
regardless of anything else that may be present in the solution. Some of the other substances that are
placed in water solution may become involved with either the hydrogen or hydroxide ions and alter the
equilibrium state. In all cases, as long as the temperature is 25°C, the equilibrium, H<iOn\ ^ H + + OH~ ,
will shift and maintain the equilibrium constant, K w , at exactly 1 x 1CT 14 .

For example, a sample of pure water at 25°C, has a [H + ] equal to 1 x 10~ 7 M and a [OH"] = 1 x 10^ 7 M.
The K w for this solution, of course, will be 1 x 10~ 14 . Suppose some HCl gas is added to this solution so
that the H + concentration increases. This is a stress to the equilibrium condition. Since the concentration
of a product is increased, the reverse reaction rate will increase and the equilibrium will shift toward the

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reactants. The concentrations of both ions will be reduced until equilibrium is re-established. If the final
[H + ] — 1 X 10~ 4 M, we can calculate the [0H~] because we know that the product of [H + ] and [0H~] at
equilibrium is always 1 x 10 -14 .

K w = [H + ][OH-] = lxl(T 14

. lxl(T 14 lxlO~ 14 in

OH-} = r - = r = 1 x 10" 10 M

L J [H+] lxlO" 4

Suppose, on the other hand, something is added to the solution that reduces the hydrogen ion concentration.
As soon as the hydrogen ion concentration begins to decrease, the reverse rate decreases and the forward
rate will shift the equilibrium toward the products. The concentrations of both ions will be increased until
equilibrium is re-established. If the final hydrogen ion concentration is 1 x 10~ 12 M, we can calculate the
final hydroxide ion concentration.

Kw —

[H + ][OH-] = lxl(T 14

, lxlO~ 14 lxlO -14

\OH~\ = r n = tt = 1 x 10~2 M

L J [H+] lxlO- 12

Using the K2 expression and our knowledge of the K2 value, anytime we know either the [H + ] or the [OH ]
in a water solution, we can always calculate the other one.

Sample question: What would be the [H + ] for a grapefruit found to have a [OH~] of 1.26 x 1CT 11 mol/L?
Is the solution acidic, basic, or neutral?

Solution:

K w = [H + ][OH~] = 1.00 x 10 -14

r ,, 1.00 xlO -14 1.00 xlO -14

\H + ] = — : : — = rr = 7.94 X 10" 4 M

L J [OH-] 1.26 xlO- 11

The pH Scale
Introduction to pH

The pH scale developed by S0rensen is a logarithmic scale. Not only is the pH scale a logarithmic scale
but by defining the pH as the negative log of the hydrogen ion concentration, the numbers on the scale
get smaller as the hydrogen ion concentration gets larger. For example, pH = 1 is a stronger acid than
pH = 2 and, it is stronger by a factor of 10. A solution whose pH = 1 has a hydrogen ion concentration
of 0.10 M while a solution whose pH = 2 has a hydrogen ion concentration of 0.010 M. You should note
the relationship between 0.10 and 0.010, 0.10 is 10 times 0.010. This is a very important point when using
the pH scale.

As was pointed out earlier, when the [H^O + ] is greater than 1 x 10~ 7 , a solution is considered to be an acid.
However, there are a great number of possibilities when you consider that an acid can have a hydrogen ion
concentration that is anything greater than 1 x 10 -7 . The same can be seen to be true for bases - just the
opposite direction (Figure 17.3).

There are a couple of important points to note here. First, all the numbers on the pH scale represent
numbers with negative exponents and exponential numbers with negative exponents are small numbers;
so all of these numbers represent [H + ] that are less than one. Second, there are acids and bases whose H +
concentrations do not fit within this range and therefore, not all acid or base strengths can be represented
on the pH scale.

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The pH scale found in Figure 17.3 shows that acidic solutions have a pH within the range of up to but
not including 7. The closer the pH is to the greater the concentration of H^O + ions and therefore the
more acidic the solution. The basic solutions have a pH with the range from 7 to 14 (Table 17.4). The
closer the pH is to 14, the higher the concentration of OH' ion and the stronger the base. For 25°C, a pH
of 7 is neutral which means that [H 3 + ] = [OH~] = 1 x 1(T 7 M.

Table 17.4:

pH level Solution

pH < 7 acidic

pH = 7 neutral

pH > 7 basic

Acid Base

Neutral

1 2 3 4 5 6 78 9 10 11 12 13 14

Figure 17.3: The pH Scale.

pH = -log [H+]

S0rensen's idea that the pH would be a simpler number to deal with in terms of discussing acidity level
led him to a formula that relates pH and \H + \. This formula is:

pH = -log [H + ]

where p = -log and H refers to the hydrogen ion concentration. The p from pH comes from the German
word potenz meaning power or the exponent of. In this case the exponent is 10. Therefore, [H + ] = 10 _pW .

When the [H+] = 0.01 mol/L, the pH will be

pH = -log (0.01)
pH = -log (1 x 10" 2 )
pH = 2

Since we are talking about negative logarithms (-log), the more hydrogen ions that are in solution, the
more acidic the solution and the lower the pH.

The Mathematics of pH

pH measures the level of acidity in solution within a certain range and has the definition, pH = -log [H + ].
pH is a logarithmic scale which means that a difference of 1 in pH units indicates a difference of a factor
of 10 in the hydrogen ion concentrations. If we have the pH of a solution and are asked to find the [H + ],
the formula for pH can be converted to a formula for [H + ] by taking the inverse log of both sides of the
equation. That process yields:

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[H+] = 10

Example 1: Determine the pH of a solution that has a [H + ] = 1 X 10 8 .
Solution:

pH = -log [H + ] = -log (1 x 10~ 8 )

The log of 1 is and the log of 10~ 8 is -8.

pH = -(0 - 8) = 8

Sample question 2: Calculate the [H + ] given that the pH is 4.
Solution:

[H+] = 10~ p//

[//+] = 10~ 4

[H + ] = 1 x 10~ 4 mol/L

Sometimes you will need to use a calculator.

Sample question 3: Calculate the pH of saliva with [H + ] = 1.58 x 10~ 6 mol/L.

Solution:

pH = -log [H + ] = -log (1.58 x 10~ 6 )
pH = 5.8

Sample Question 4: Fill in the rest of the Table 17.5.

pH = -log [H + ]

Table 17.5:

[H+] (mol/L) -log [H+] pH

0.1 1.00 1.00

0.2 0.70 0.70

1.00 xl0~ 5 ? ?

? ? 6.00

0.065 ? ?

? ? 9.00

Solution:

pH = -log [#+]

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Table 17.6:

[H+] (mol/L) -log [H+] pH

0.1 1.00 1.00

0.2 0.70 0.70

1.00 xl0~ 5 5.00 5.00

1.00 xl0~ 6 6.00 6.00

0.065 1.19 1.19

1.00 xl0~ 9 9.00 9.00

We said that the pH scale was one that showed the pH becoming lower as the strength of the acid becomes
larger. Let's think about this for a second. Stomach acid is HCl, a strong acid. Strong acids are powerful,
we can assume because they completely ionize and therefore would have all of their H + ions present in the
solution when the reaction is complete. Vinegar (that we may put on our salad, cook with, make those
neat science fair volcanoes with!) is a weak acid. It only partially ionizes and only allows some of its H +
ions to come into solution. Therefore, the pH of HCl, according to this observation would be lower than
that of vinegar. Look at Figure 2. What does the pH scale diagram tell us about the pH of 0.1 M HCl
and 5% vinegar? Sure enough, the pH for HCl is 1.0 and that of 5% vinegar is around 2.8.

Have you ever cut an onion and had your eyes water up? This is because of a compound with the formula
C^HqOS that is found in onions. When you cut the onion, a variety of reactions occur that release a
gas. This gas can diffuse into the air and eventfully mix with the water found in your eyes to produce a
dilute solution of sulfuric acid. This is what irritates your eyes and causes them to water. There are many
common examples of acids and bases in our everyday lives. Look at the pH scale (Figure 17.4) to see how
these common examples relate in terms of their pH.

Soda Pop

Pure Water Soa P

NaOH

HCl Vinegar

. I Tomatoes Milk Eqqs Detergent '

L J i i u ill

01 2 3.45 6 7T 89 10 12 13 14

Blood . T

I Ammonia

Lemon Juice Milk of Magnesia

Orange Juice

Figure 17.4: pH Scale for Common Substances.

You can see that pH definitely does play a role in our everyday lives as we come in contact with many
substances that have varying degrees of acidity. We may not necessarily think about the pH of eggs as we
eat them or think about how acidic orange juice is when we enjoy a glass, but maybe next time we have
breakfast we may ponder for a second about the work of S0rensen ... just maybe!

Summary

• Water dissociates to a very slight degree according to the equation ^Ora ^=> [H + ] + [OH~].
. In pure water at 25°C, [H+] = [OH~] = 1.00 x 10~ 7 M.

• The equilibrium constant for the dissociation of water, called K w , at 25°C is equal to 1.00 x 10 -14 .

• The definition of pH is pH = -log [H + ].

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Review Questions

1. Why is it necessary to balance the chemical equation before determining the [H + ] or [0H~\ for strong
acids and bases?

2. Why can't you determine the [H + ] or [OH'] for weak acids and bases the same way you can determine
the [H + ] or [0H~] for strong acids and bases?

3. What is the [H + ] ion concentration in a solution of 0.350 mol/L H2SO4?

(a) 0.175 mol/L

(b) 0.350 mol/L

(d) 1.42 x lO' 14 mol/L

4. A solution has a pH of 6.54. What is the concentration of hydronium ions in the solution?

(a) 2.88 x 10" 7 mol/L

(b) 3.46 x 10" 8 mol/L

(d) 7.46 mol/L

5. A solution has a pH of 3.34. What is the concentration of hydroxide ions in the solution?

(a) 4.57 x 10~ 4 mol/L

(b) 2.19 x 10 -11 mol/L

(d) 10.66 mol/L

6. A solution contains 4.33 x 10~ 8 M hydroxide ions. What is the pH of the solution?

(a) 4.33

(b) 6.64

(d) 9.67

7. Fill in the Table 17.7 and rank the solutions in terms of increasing acidity.

Table 17.7:

Solutions [H+] (mol/L) -log [H + ] pH

A 0.25 0.60 0.60

B ? 2.90 ?

C 1.25 xl0~ 8 ? ?

D 0.45 xl0~ 3 ? ?

E ? 1.26 ?

8. A bottle of calcium hydroxide is found in the lab with a label reading: 0.014 mol/L.

(a) What are the concentrations of all of the ions present in the solution?

(b) What is the pH of the solution?

9. It has long been advocated that red wine is good for the heart. However, wine is also considered to
be an acidic compound. Determine the concentration of hydronium ions in wine with pH3.81.

10. The diagram that follows represents a weak acid before ionization and when the reaction comes to
equilibrium. If the acid were weaker than the one represented in the diagram, how would the diagram
change? Draw a new diagram to represent your answer. If the acid were a strong acid, how would

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the diagram change? Draw a new diagram to represent your answer.

100
90 f-
80 -|~

■ Weak Acid

■ H* Ion
□ Anion

Equilibrium

Further Reading / Supplemental Links

• http://en.wikipedia.org

Vocabulary

pH scale A scale measuring the [H + ] with values from to 14.

pH = -log [H + ] - Formula used to calculate the power of the hydronium ion.

autoionization Autoionization is when the same reactant acts as both the acid and the base.

ion product constant for water K w , is the product of the hydronium ion and the hydroxide ion con-
centrations in the autoionization of water.

17.6 Weak Acid/Base Equilibria

Lesson Objectives

• Define weak acids and weak bases in terms of equilibrium.

• Define K a and Kb-

• Use K a and K b to determine acid and base strength.

• Use K a and Kb in acid/base equilibrium problems.

Introduction

Acid and base equilibria are another form of equilibrium reactions that deserve special mention. In prior
learning we had discussed that weak acids and bases do not completely ionize or dissociate in solution.
Thus, their reactions also possess an equilibrium state where their forward reaction rates equal their reverse
reaction rates. Since this is the case, equilibrium constant expressions and equilibrium calculations can
be completed for these reactions as was done for all other equilibria. When there are a group of reactions
that are similar in appearance and function, the equilibrium constant for those reactions are often given a
subscript denoting them as belonging to the group. Such was the case for solubility product constants which
were given the designation K sp . The dissociation of weak acids are also a special group of reactions and
their equilibrium constant is designated K a . The equilibrium constants for the dissociation of weak bases
are designated Kb- In this section, we will focus our attention to a special case of equilibrium reactions,
the acid - base equilibria.

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Weak Acids and Weak Bases are Equilibrium Systems

Strong acids are denned as acids that completely ionize in water solution; strong bases as those that
completely dissociate in water solution. Look at Equation 1 and Equation 2 for an example of a strong
acid ionization reaction and a strong base dissociation reaction, respectively.

HClfa) -» H + {aq) + Cl~ {aq) (Equation 1)

NaOH^ -> Na + ( a? ) + OH~ ^ (Equation 2)

Weak acids and bases, however, do not completely ionize or dissociate in solution. In the chapter on weak
acids and bases, we learned that solutions of weak acids and bases ionize or dissociate much less than
100%. Ammonia is a weak base, for example. Look at Equation 3, the dissociation reaction for ammonia.

NH 3{aq) + H 2 {L) £; NH 4 + {aq) + OH~ {aq) (Equation 3)

Ammonia has an equilibrium constant, Kb, of 1.8 x 10~ 5 . This value is less than 1. This means that the
equilibrium position favors the reactants and that there is not very much reaction of ammonia to form the
protonated ammonium ion. Said another way, at equilibrium, there are many more ammonia molecules
present in solution than there are ammonium ions or hydroxide ions.

The same is true for weak acids. Vinegar is a common weak acid and represents an approximately 5%
acetic acid solution. Equation 4 shows the ionization of acetic acid or vinegar.

HC 2 H 3 2 (a q ) ^> H+ {a q ) + C 2 H 3 2 ~ ( fl(? ) (Equation 4)

The equilibrium constant, K a , for acetic acid is 1.8 x 10 _o . Again, this value is small indicating that the
equilibrium position lies more to the left than to the right. In other words, there are more acetic acid
molecules at equilibrium than there are acetate ions or hydronium ions. Since K a is less than 1, we know
that [HC 2 H 3 2 ] > [C 2 H 3 2 ~] or [H + ] at equilibrium.

Sample question 1: Put the following acids in order of decreasing acid strength. Write equilibrium reactions
for each. Remember K a is the equilibrium constant for the acid ionization which increases with increasing
acid strength (or decreasing pH).

Formic acid, HCOOH, K a = 6.3 x 10~ 4

Phosphoric acid, H 3 P0 4 ,K a = 7.2 x 10~ 3

Oxalic Acid, HCOOCOOH,K a = 5.6 x 10~ 2

Arsenic acid, H 3 AsOi,K a = 6.0 x 10~ 3 .

Solution:

Order of decreasing acid strength: Oxalic acid > Phosphoric acid > Arsenic acid > Formic acid

Equilibrium Reactions:

1. Oxalic acid: HCOOCOOH {aq) ±? H + {aq) + HCOOCOO {aq)

2. Phosphoric acid: H 3 PO A ^ aq ) i? H + {aq) + H 2 PO/C \ aq)

3. Arsenic acid: H 3 AsO i{aq) i? H + \ aq) + H 2 As04T{ aq )

4. Formic acid: HCOOH (aq) ±* H + {aq) + COOH \ aq)

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Ka and Kb

The pH of solutions of strong acids and strong bases can be calculated simply by knowing the concentration
of the strong acid and, of course, knowing that these acids and bases dissociate in solution 100%. Consider
a solution that is 0.010 M HC1. HCl is a strong acid and therefore, the acid molecules dissociate 100% and
this 0.010 M solution of HCl will be 0.010 M in H + ion and 0.010 M in Ct ion. Plugging the value of the
hydrogen ion concentration into the pH formula, we can determine that this solution has a pH = 2.

Consider a solution that is 0.0010 M NaOH. NaOH is a strong base and therefore, this solution will be
0.0010 M in sodium ions and also in hydroxide ions. Since the solution is 1.0 X 10 -3 M in hydroxide ions,
it will be 1.0 X 10~ n M in hydrogen ions. Therefore, this solution will have a pH = 11.

A diatomic strong acid such as H 2 S0 4 is only slightly more complicated. Suppose we wish to determine
the pH of a 0.00010 M solution of H 2 S0 4 .

H 2 S0 4(aq) -»2 H + + S0 4 2 ~

Since the original solution was 0.00010 M in the acid molecules and sulfuric acid is a strong acid, then the
100% dissociation will produce a solution that contains [H + ] = 0.00020 M. Substituting this hydrogen ion
concentration in the pH formula yields:

pH = -log (2.0 x 10~ 4 )

pH = -(0.30 - 4) = -(-3.7) = 3.7

Let's now consider the process for finding the pH of weak acids and bases. In these cases, you need more
information than you need for strong acids and bases. Not only do you need to know the concentration of
the original acid or base solution, but you also must know the K a or K/,.

Suppose we wish to know the pH of a 1.0 M solution of ascorbic acid, H2C%H^O^i a ^\, whose K a = 1.8 x 10 -5 .

^C^H^O^a^ £5 H ( fl? ) + HCqHqOq {aq) K a = 1.8 X 10

The K a for this reaction would be written

[H+][HC 6 H 6 6 -]

K a

[H 2 C q H & Oq]

To find the hydrogen ion concentration from this K a expression and the original concentration of the acid,
we need a little algebra. We need to assign some variables.

Let the molarity of the acid that dissociated be represented by x. Then the molarity of the hydrogen
ions and ascorbate ions in solution will also be represented by x. The molarity of the undissociated acid
remaining would be represented by 1.0 — x. We can now substitute these variables into the K a expression
and set it equal to the given K a value.

(X)(X) r

K a = , - \ = 1.8 X 10" 5 Expression 1

(1.0 - x)

When this equation is simplified, we find that it is a quadratic equation . . . which, of course, can be solved
by the quadratic formula.

x 2 + (1.8xl0~ 5 )x-1.8xl0~ 5 = and x = 4.2 x 10~ 3 M

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However, there is a shortcut available to solve this problem that simplifies the math greatly. It involves
significant figures and adding or subtracting extremely small numbers from large numbers. If you are
working to 3 significant figures and you are required to subtract 0.00005 from 1.00, when you carry out the
subtraction and round to 3 significant figures, you discover that you get the original number before you
subtracted.

1.00 - 0.00005 = 0.99995 which to 3 significant figures is 1.00

In the problem we solved above about ascorbic acid, the K a value is very small, 1.8 x 10~ 5 . This indicates
that the amount of ascorbic acid that dissociates, represented by x, is tiny. When we assigned the variables
in that problem, we see that the molarity of ascorbic acid remaining after dissociation is represented by
1.0 - x. Since this x is very tiny, the result of this subtraction will be 1.0 M. Therefore, the K a expression
from above,

K a = / A '— = 1.8 X 10~ 5 Expression 1

(1.0 - x)

can quite safely be written as

(x)(x) K „

K a = jfftf- = 1-8 x 10~ 5 Expression 2

You should notice that the variable 1.0 - x has been replaced simply by 1.0. We are assuming that the x
is so small, it will not alter the value 1.0 when we subtract and then round to proper significant figures.
This is a safe assumption when the value of the K a is very small (less than 1 x 10~ 3 ).

When we solve this second expression

(x)(x) K

VV = 1-8 xlO" 5
(1.0)

x 2 = 1.8xl0~ 5 and

x = 4.2 x 10~ 3 M

We no longer need to use the quadratic formula and you must note that the answer to the proper number
of significant figures is exactly the same as when Expression 1 was solved by the quadratic formula.

Let's go through another of these problems finding the pH of a weak acid given the initial concentration
of the acid and its K a . This time we will use a hypothetical weak acid, 0.10 M HA, whose K a = 4.0 x 10 -7 .

HA (aq) ±+H + + A~ K a = 4.0 x 10~ 7

[H+M-] -

K —- - --40x1(1"'

K a - [ha] - 4.U X 1U

Let x represent the molarity of HA that dissociates, then [H + ] = [A~] = x, and [HA] = 0.10 - x.

K a = r ; = 4.0 X 10" 7

[0.10 - x]

Once again, since the K a value is very small, the x subtracted in the denominator can be neglected and
the equation becomes

Ixl Ixl 7

! " ' -4.0xl0 -7

[0.10]

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so, x 2 = 4.0 x 1CT 8 and x = 2.0 x 10~ 4 M.

Therefore, the hydrogen ion concentration in this solution is 2.0 X 10~ 4 M and substituting this value into
the pH formula yields

pH = -log (2.0 x 10~ 4 ) = -(0.3 - 4) = -(-3.7) = 3.7

The same process is used for weak bases. There is one additional step when working with weak bases because
once the hydroxide ion concentration is determined, you must then find the hydrogen ion concentration
before substituting into the pH formula. Kj, represents the equilibrium constant for the dissociation of a
weak base. Look at the dissociation of dimethylamine (a weak base used in making detergents). We will
calculate the pH of a 1.0 M solution of this weak base.

(CH 3 ) 2 NH {aq) + H 2 {L) fc> (CH 3 ) 2 NH 2 + {aq) + OH~ {aq) K b = 5.9 x 10" 4

WW 7

[(CH 3 ) 2 NH]

Allowing x to represent the molarity of {CH 3 ) 2 NH that dissociates results in [{CH 3 ) 2 NH 2 + \ = x and
[OH—] = x. The molarity of undissociated (CH 3 ) 2 NH will be 1.0 - x.

Substituting the variables into the Kb expression yields

Ijcl be] 7

1 Jl J -4.0xl0~ 7

[1.0 -x]

and neglecting the x subtracted in the denominator because it is beyond the significant figures of the
problem yields

\x] be] 7

! ' ! ' - 4.0 x 10" 7 .

[1.0]

Therefore, x 2 = 4.0 x 10" 7 and i = 6.3x 10~ 4 M.

Now that we know the hydroxide ion concentration in the solution, we calculate the hydrogen ion concen-
tration by dividing the [OH~] into the K w . This will yield [H + ] = 1.6 x 10~ n M. The final step is to plug
the hydrogen ion concentration into the pH formula.

pH = -log (1.6 x lO" 11 ) = 10.8

Sample question 2: Acetic acid is mixed with water to form a 0.10 mol/L HC 2 H 3 2 t aq \ solution at 25° C. If
the equilibrium concentrations of H 3 + r aq \ and C 2 H 3 2 ~ i aq \ are both 1.34xl0~ 3 mol/L and the equilibrium
concentration of HC 2 H 3 2 ^ is 0.0999 mol/L, determine the

(a) K a and

(b) the pH of the solution at equilibrium.
Solution:

HC 2 H 3 2 ( aq ) <=> # + ( a? ) + C 2 H 3 2 ( a? )

( a \ K ._ [H+][C 2 H 3 Q 2 -] _ (1.34xl0- 3 )(1.34xl0- 3 ) _ i on v i n-5
W ^" - IHC 2 H 3 2 ] ~ (0.0999) ~ 1 " OU A 1U

(b)

pH = -log [H + ]

pH = -log (1.34 x 10" 3 )

pH = 2.87

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Lesson Summary

• Weak acids only partially ionize in solution and therefore represent equilibrium reactions. Weak bases
only partially dissociate in solution and therefore represent equilibrium reactions. K a represents the
equilibrium constant for the ionization of a weak acid.

• Kb represents the equilibrium constant for the dissociation of a weak base. Equilibrium calculations
are the same for weak acids and bases as they were for all other equilibrium reactions.

Review Questions

1. What makes weak acids and bases a special case for equilibrium reactions?

2. What do the constants K a and Kb represent?

3. Oxalic acid is a weak acid. Its ionization reaction is represented below. H2C2O Uiaq) + H 2 On\ ^>
H%0 + t aq \ +HC 2 0&~ Uq) Which of the following best represents the acid ionization constant expression,
K f

"■a ■

(a) K a

(b) K a

(d) K a

lH 3 0+][HC 2 0- 4 ]
[H 2 C 2 4 ][H 2 0]
[H 2 C 2 Q 4 }[H 2 0]
[H 3 0+][HC 2 0- 4 ]
[H 3 Q+][HC 2 Q- 4 ]
W2C2O4]
[H 2 C 2 Q 4 ]

[H 3 0+][HC 2 0~ 4 ]

4. Choose the weakest acid from the list below.

(a) HN0 2{aq) ;K a = 5.6 X 10" 3

(b) HF {aq) ;K a = 6.6xlO- 4

(d) HCOOH {aq) ;K a = 1.8 X 1CT 4

5. Choose one of the following reactions that would best represent a reaction that has an equilibrium
constant best described as a base dissociation constant, Kb.

(a) H 2 P0 4 ~ {aq) + H 2 {L) ±5 HP0 4 -\ aq) + H 3 0+ {aq)

(b) NH A + {aq) + H 2 {L) U NH Haq) + H 3 0+ {aq)

(d) F- {aq) + H 2 {L) <^ HF {aq) + OH- {aq)

6. A 0.150 mol/L solution of a weak acid having the general formula HA is 15.0% ionized in aqueous
solution. Which expression best represents the calculation of the acid ionization constant K a for this
acid?

/ s K _ (0.150)(0.150)

W ^ a ~ (0.150)

/, n „ _ (0.0225)(0.0225)

[0) &a - ( 0128 )

7. Put the following bases in order of increasing base strength. Write equilibrium reactions for each,
ethanolamine (HOCH 2 CH 2 NH 2 ),K h = 3.2 X 10~ 5 , piperidine (C 5 H w NH),K h = 1.3 X 10~ 3 , triethy-
lamine {{CH 3 CH 2 ) 3 N),K b = 5.2 x 10~ 4 , and ethylenediamine (H 2 NCH 2 CH 2 NH 2 ),K h = 8.5 x 10~ 5 .

Further Reading / Supplemental Links

• http://en.wikipedia.org/wiki

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Vocabulary

acid ionization constant K a represents the equilibrium constant for the ionization of a weak acid.
base dissociation constant Kb represents the equilibrium constant for the dissociation of a weak base.

17.7 Br0nsted Lowry Acids-Bases

Lesson Objectives

• Define a Br0nsted-Lowry acid and base.

• Identify Br0nsted-Lowry acids and bases from balanced chemical equations.

• Define conjugate acid and conjugate base.

• Identify conjugate acids-bases in balanced chemical equations.

• Identify the strength of the conjugate acids and bases from strengths of the acids and bases.

Introduction

Arrhenius provided chemistry with the first definition of acids and bases but like a lot of theories, this
model tended to be refined over time. This is exactly what happened in the area of acid/base chemistry.
Two chemists, named Br0nsted and Lowry, working on similar experiments as Arrhenius, derived a more
generalized definition for acids and bases that we use in conjunction with the Arrhenius theory. The
Br0nsted-Lowry theory is the focus of this lesson. As the Br0nsted-Lowry definition unfolded, the number
of acids and bases that were able to fit into each category increased. Thus, the definitions were broader
for each.

Br0nsted-Lowry Acids and Bases
Br0nsted-Lowry Definitions

Arrhenius made great in-roads into the understanding of acids and bases and how they behaved in chemical
reactions. Br0nsted and Lowry slightly altered the Arrhenius definition and greatly enlarged the number
of compounds that qualify as bases. The Br0nsted-Lowry theory defines an acid as a substance that is a
proton donor and a base as a proton accepter. Look at equation 1 in which hydrochloric acid is reacting
with water:

HCl (s) + H 2°(L) -> H 3 + ^ m]) + Cr^ aq) (Equation 1)

What is happening is that the acid, HCl is losing a H + ion to form CV and the H2O is gaining a H + ion
to form H$0 + . The Br0nsted-Lowry concept of acids and bases states that the acid donates a proton
and the base accepts a proton. Therefore HCl is donating an H + ion to H2O to form Cl~ and the H2O is
accepting an H + ion from HCl to form H%0 + ; HCl is acting as the acid and H20{aq) is acting as the base.
Look at equation 2 to see another example of the Bronsted-Lowry Theory in action.

H 2 POf {aq) + OH- (aq) U HPOi 2 - {aq) + H 2 (L) (Equation 2)

We can see that again, the equation shows H2PO4 is donating a proton to OH to form HPO^~ and OH
is accepting the proton to form H2O. Thus H2PO^{aq) is acting as the acid and OH~{aq) is acting as the
base.

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Sample question: Identify the Br0nsted Lowry Acids and Bases from each of the following equations:

(a) HC 2 H 3 2 ^ + H 2 0( L ) <=> C 2 H 3 2 ^ + H 3 + ( aq )

(b) HCN {aq) + H 2 {L) £5 CN- {aq) + H 3 0+ {aq)
Solution:

HC 2 H 3 2 ( aq) + #2<?(L) ±+ C 2 H 3 2 ( a? ) + // 3 ( aq )

Acid base

(b)

#CAr (a<?) + H 2 (L) i? CN- (aq) + H 3 + {aq)
Acid base

Br0nsted— Lowry Acids/Bases Definitions Includes More Compounds

If you think about the definition of an Arrhenius acid, it includes substances such as HCl, HN0 3 , HC 2 H 3 2 ,
in essence all substances that contain H + ions. This is because according to Arrhenius, the acid ionizes in
water to produce H + ions. This definition limits what can fit under the umbrella of the definition of acid.
The Br0nsted-Lowry definition of the acid is broader in that it defines the acid as a proton donor. With
this broader definition there is the ability to include more compounds in the category of acid. Consider
the reaction Equation 3:

HSO A ~ {aq) + OH~( aq) i? H 2 {L) +S0 4 2 ~( aq ) (Equation 3)

It needs to be pointed out that if a substance is an acid in the Arrhenius definition, it will be an acid in
the Br0nsted-Lowry definition. However, the reverse is not true. Nor, do Br0nsted-Lowry acids now have
the properties of an Arrhenius acid. The same is true for bases. In equation 3, the hydroxide ion, OH~, is
both an Arrhenius base and in the reaction a Br0nsted-Lowry base. In other words, the Br0nsted-Lowry
definition can be viewed as an extension to the Arrhenius definition rather than a replacement for it.

May or May Not be in Water Solution

With the Arrhenius theory, one aspect that is consistent, is that water was part of the equation. Arrhenius
said that an acid must produce H + ions in a water solution. Therefore, according to Arrhenius, the
following equation would be representative of an Arrhenius acid.

HCl {g) + H 2 {L) -> H 3 + {aq) + OH- {aq) (Equation 4)

If you look at all of the equations we have used thus far for Arrhenius, and even the definitions in Arrhenius
theory, one commonality shows through: water must be in the equation. Here is where the Br0nsted-Lowry
definition again varies from Arrhenius theory. Look at the equation below.

NH 3{aq) + NH 3{aq) ±5 NH 4 + {aq) + NH 2 - (aq) (Equation 5)

When you look at Equation 5, the first NH 3 molecule is accepting a proton to form NH^ and is therefore
a Br0nsted-Lowry base, the second NH 3 t aq \ molecule is donating a proton to form NH 2 ~ and is therefore a

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Br0nsted-Lowry acid. Ammonia molecules, however, do not donate hydrogen ions in water and therefore,
do not qualify as Arrhenius acids. The Br0nsted-Lowry theory has provided a broader theory for acid-base
chemistry.

It should be noted that NH 3 is an example of an amphoteric species. Amphoteric species are those that
in different situations can act as either an acid or a base. That is, in some circumstances, they donate
a proton and in other circumstances, they accept a proton. Here in Equation 5, NH 3 is doing just this.
Water is also an amphoteric species.

Demonstrations explain pH and how it is measured, and the important role of acids and bases. The Proton
in Chemistry (http : //www . learner . org/vod/vod_window . html?pid=808)

Acid-Base Conjugate Pairs Definition

There is one more aspect of the Br0nsted-Lowry theory that was a significant breakthrough to acid-base
chemistry. Br0nsted and Lowry said that in acid-base reactions, there are actually pairs of acids and bases
in the reaction itself. According to Bronsted-Lowry, for every acid there is a conjugate base associated
with that acid. The conjugate base is the result of the acid losing (or donating) a proton. Therefore, if
you look below, you can see on the left, the acid and on the right the conjugate base.

Acid Conjugate Base

HCI -> CI

HBr -> Br

HN0 3 -> N0 3

HC 2 H 3 2 -> C 2 H 3 0~ 2

Notice that the difference between the acid and its conjugate base is simply a difference of a proton. The
conjugate base has one less proton (or H + ). Conjugate acids work the same way. For every base in the
acid-base reaction, there must be a corresponding conjugate acid. The conjugate acid is the result of
the base gaining (or accepting) the proton. Look below to see the difference between the base and the
corresponding conjugate acid.

Base Conjugate Acid

NH 3 -> NH+

OH -> HOH

H 2 -> H 3 +

COf -> HCOl

Now that we know what a conjugate acid and base is, let's now try to identify them in acid-base reactions.
Look at Equation 6 below, a reaction between acetic acid and water.

C 2 H 3 2 ~( aq ) + H 3 + {aq) (Equation 6)

HC 2 H 3 2 (aq)

H 2 0( L ) -

Acetic Acid

Water

Step

1: Identify the acid and base.

HC 2 H 3 2 ( aq )

+ H 2 {L)

Acid

Base

<=> C 2 H 3 2 ( ag ) + H 3 ( aq)

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Step 2: Identify the conjugate acid and base on the product side of the equation. Look at the product side
to see what product has gained a proton (this is the conjugate base) and which product has lost a proton
(this is the conjugate acid).

HC2H- i 2 ( a q) + #2<2(L) ^> C2H3O2 ( a q) + H 3 ( m/ )

Acid Base Conjugate Base Conjugate Acid

As a result, the conjugate acid/base pairs are HC<iHz02{aq) j CiHj,0<i~ (aq) and H2O Vn / H 3 + (aq)-
Now you try. Identify the conjugate acid-base pairs in the following equation.

CH 3 NH 2{aq) + HCIO {aq) £5 ClO- {aq) + CH 3 NH 3 + {aq) (Equation 7)

First, identify the acid and base on the reactant side.
Step 1:

CH 3 NH 2{aq) + HClO {aq) i? ClO~ (aq) + CH 3 NH 3 + {aq)

Base Acid

Then identify the conjugate acid and base in the products.
Step 2:

CH 3 NH 2{aq) + HClO {aq) t; ClO~ {aq) + CH 3 NH 3 + (aq)

Base Acid Conjugate Base Conjugate Acid

Hence, the conjugate acid/base pairs are CH 3 NH2/CH 3 NH^ and HCIO / ClO~ .

Sample Problem: Identify the conjugate acid-base conjugate pairs in each of the following equations:

(a) NH 3{aq) + HCN {aq) ±+ NH 4 + {aq) + CN~ (aq)

(b) C0 3 2 - {aq) + H 2 {L) U HC0 3 - {aq) + OH- {aq)
Solution:

(a)NH 3 /NH+ and HCN/CN-
(b)C0 3 2 -/HC0 3 - and H 2 0/OH-

The Strength of Conjugate Acids and Bases

The definition of a Br0nsted-Lowry acid is that it donates a proton (hydrogen ion) . In order to be a strong
Br0nsted-Lowry acid, it must donate the proton readily. That is, the bond between the hydrogen ion
and the cation is weak so that the proton is released easily. The definition of a Bronsted-Lowry base is
that it accepts a proton. In order to be a strong base, it must take on that proton readily. Therefore, a
Br0nsted-Lowry strong base would hold onto the proton tightly.

If we consider a strong acid, such as HCl, we know that the acid releases its proton readily . . . does that
tell us anything about the Cl~ ion as a base? If the chloride ion does not hold onto the proton tightly, we
know HCl will be a strong acid, and therefore, because the chloride ion releases the proton easily, it cannot
be a strong base. The same behavior that qualifies an anion attached to a hydrogen ion as a strong acid
also qualifies the anion alone as a weak base.

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The reverse of this situation would also be true. Consider a weak acid such as acetic acid, HC2H%02- This
is a weak acid because the acetate ion does not release the hydrogen ion readily; it holds it tightly, hence
only ionizes to a slight degree. That tells us that the acetate ion takes on a proton readily and holds it
and that qualifies the acetate ion as a strong base.

Strong acids produce conjugate bases that are weak bases and weak acids produce conjugate
bases that are strong bases.

In this same way, a strong base such as OH~ ion takes on hydrogen ions readily and holds them tightly.
Therefore, the conjugate base of OH~ ion, which is H2O, will be a weak acid. A weak base such as ammonia,
NH3, does not take on a hydrogen ion readily and that causes it to be a weak base. Therefore, the conjugate
acid of NH3, NH4 + , will be a strong acid.

Strong bases produce conjugate acids that are weak acids and weak bases produce conjugate
acids that are strong acids.

See Table 17.8, which conjugate bases would be strong and which would be weak?

Table 17.8: Strong Acids

Name

Symbol

Conjugate Base

Strength of Conjugate

Base

Weak

N0 3 -

Weak

ClO A -

Weak

HSO4-

Weak

Hydrochloric Acid

HCl

Hydrobromic Acid

HBr

Hydroiodic Acid

Nitric Acid

HN0 3

Perchloric Acid

HClO A

Sulfuric Acid

H2SO4

All of the strong acids will have weak conjugate bases. All other acids are weak and will therefore have
conjugate bases that are strong.

The stronger the acid, the weaker the conjugate base. This means that the conjugate bases will follow a
similar trend as the acids only in the reverse. Look at Table 2. See how the acid strength trend compares
with the conjugate base strength trend.

Table 2: Strong Acids and Their Conjugate Bases

Increasing

Acid
Strength

Strong Acid

Conjugate Base

HCl

CI"

HBr

Br"

HNOj

NOj

HCIO a

cio;

H z S0 4

hso;

Increasing

Base
Strength

As we have seen with the strong acids-weak conjugate bases and weak acids-strong conjugate bases, the
same concept can be applied to bases. Strong bases have weak conjugate acids and weak bases have strong
conjugate acids. Which conjugate acids would be strong and which would be weak?

Only the OH— ion represents a strong base and therefore only H2O represents a weak conjugate acid. All
of the other bases are weak. Therefore all of the other conjugate acids are strong. The strong bases are
LiOH,NaOH,KOH,RbOH,CsOH,Ca(OH) 2 ,Sr(OH)2, and Ba(OH) 2 - All of these strong bases contain OH~
and therefore their weak conjugate acid will be H2O. The remaining bases are weak and so would have

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520

strong conjugate acids.

Lesson Summary

• The Br0nsted-Lowry concept of acids and bases states that the acid donates a proton and the base
accepts a proton.

• The Br0nsted-Lowry definition includes more substances because the definition is not confined to
substances containing hydroxides or to be in water. A conjugate acid is a substance that results
when a base gains (or accepts) a proton.

• A conjugate base is a substance that results when an acid loses (or donates) a proton.

• Strong acids result in weak conjugate bases when they lose a proton and weak acids result in strong
conjugate bases when they lose a proton.

• Strong bases result in weak conjugate acids when they gain a proton and weak bases result in strong
conjugate acids when they gain a proton.

Review Questions

1. What improvements did Br0nsted-Lowry make over the Arrhenius definition for acids-bases?

2. If you were to use HCl{aq) as an example, how would you compare the Arrhenius definition of an
acid to the Br0nsted-Lowry definition?

3. What is the Br0nsted-Lowry definition of an acid?

(a) a substance that donates protons

(b) a substance that accepts protons

(d) a substance that dissolves in water to form H + ions

4. If H 3 + is an acid according to the Br0nsted-Lowry theory, what is the conjugate base of this acid?

(a) H 4 2+ {aq)

(b) H+ {aq)

(d) OH~ {aq)

5. What is the conjugate base of H 2 P0 4 ^1

(a) H 3 + {aq)

(b) H 3 P04 (aq)

(d) P0 4 3 - {aq)

6. In the following reactions, which are the Br0nsted-Lowry acids? i) H 3 PO^ aq ^ + H 2 0^ ^=> HzO + ( aq ) +
H2PO A - {aq) ii) H 2 P0 4 - {aq) + H 2 {L) £> H 3 0+ (aq) + HPO A 2 ~ {aq)

(a) H 2 PO A ~,H20,HPO A 2 ~

(b) H 3 P0 4 ,H 2 0,H 2 P0 4 ~

(d) H 3 P04,H 3 + ,H 2 P0 4

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7. Label the conjugate acid-base pairs in each reaction.

(a) HC0 3 + H 2 0<=> H 2 C0 3 + OH

(b) H 2 POi + H 2 «=> H 3 + + HP0 4 2 ~

(d) #F(a ? ) + // 2 0(/) fl H 3 + {aq) + F-(aq)

8. Complete the following reactions. When done, label the conjugate acid/base pairs.

(a) Br0 3 +H 2 0^

(b) HN0 3 + H 2 ->

9. For the reactions in question 7, which are the weak conjugate bases and which are the strong conjugate
bases?

Further Reading / Supplemental Links

• http : //learner . org/resources/series61 . html

Vocabulary

Br0nsted-Lowry acid A substance that donates a proton (H + ).

Br0nsted-Lowry base A substance that accepts a proton (H + ).

amphoteric substances Substances that act as both acids and bases in reactions (i.e. NH 3 ).

conjugate acid The substance that results when a base gains (or accepts) a proton.

conjugate base The substance that results when an acid loses (or donates) a proton.

17.8 Lewis Acids and Bases

Lesson Objectives

• Define a Lewis acid and a Lewis base.

• Understand and define a coordinate covalent bond.

• Identify a Lewis acid and a base in reactions.

Introduction

Lewis is known for the shared-pair chemical bond and his work with electrons. We will be concentrating on
his work with electrons. You may recall working with Lewis electron dot diagrams from an earlier chapter.
Look at the example below for a quick refresher.

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CH 4 H:c:H

H
C

In the early 1930s, Gilbert Lewis saw the need for a more general definition for acids and bases that
involved the use of electrons. Using the work of Br0nsted and Lowry, he saw that in some cases the
acids may not have protons to donate, but may have electron pairs to donate. We will give a look at the
contribution of Lewis to the theory of acids and bases. It should be noted that most chemists today use the
Br0nsted-Lowry definitions of acids-bases and the Lewis definitions are used in more specialized situations.

Lewis Acids

Lewis defined an acid as a substance that accepts a pair of electrons from another substance. Therefore,
Lewis acids must have room in their structure to accept a pair of electrons. Remember that each central
atom can hold eight electrons. What this means is that if the atom does not have eight electrons, but six
electrons, then it can accept one more pair. Look at the structure in Figure 17.5. Notice how, in each
case, there is room to accept a pair of electrons.

The Lewis acid will accept the electron pair in order to form a bond. The bond that forms between the two
atoms will be covalent bonds. Covalent bonds are formed when electrons are shared between two atoms.

H +

The BF 2 molecule can hold This hydrogen ion can hold 2

3 electrons surrounding the electrons around the hydrogen

boron atom ... but in this atom ... but in this case has

case has only 6 electrons zero electrons around the

around the boron. Therefore. hydrogen. Therefore, this ion

this molecule can take on can take on two electrons and

two electrons and that makes that makes it a Lewis acid,
it a Lewis acid.

Figure 17.5: Examples of Lewis Acids.

Lewis Bases

Lewis defined a base as a substance that donates a pair of electrons to a substance. Lewis bases have a
lone pair of electrons that they can share with another species when they donate them as part of their
performance as a base. If we look at the example of bases in Figure 17.6, we can see that each of them
has a lone pair of electrons available to donate.

Again, since the base will share the electrons with the corresponding acid when it donates this pair, the
resulting bond that forms between the acid and the base will be a covalent bond.

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H
H:n: H: ° ;

ii\

This ammonia molecule has This hydroxide ion has four

four pairs of electrons around pairs of electrons around its

its central atom and only three central atom and only one

of the pairs are shared. of the pairs is shared.

Therefore, this molecule has Therefore, this ion has unshared

another pair of electrons that pairs of electrons available to

it can share and that makes share and that makes it a

it a Lewis base. Lewis base.

Figure 17.6: Examples of Lewis Bases.

Lewis Acid-Base Neutralization

We have said already that the Lewis acid accepts a pair of electrons from a Lewis base that donates the
pair of electrons. The resulting bond that forms is a covalent bond. In a regular covalent bond, each of
the bonded atoms contributes one of the shared pair of electrons. In certain cases, it is possible for one of
the bonded atoms to contribute both of the shared electrons and the other atom contributes no electrons.
Coordinate covalent bond is the term given to bonds formed when both electrons come from the same
atom. Look at the equation below.

H IF H F

H : N : + B : F ^- H : N : B : F

H F H F

This is a Lewis acid combining with a Lewis base and therefore, this reaction represents a Lewis acid-base
neutralization reaction. Since both of the electrons that participate in the bond between TV and B have
come from the Lewis base NH 3 , the N B bond is a coordinate covalent bond.

Sample question: Identify the Lewis acid and Lewis base in each of the following reactions. Then write
the reactions to show that these reactions involve coordinate covalent bonds.

(a) Ag + {aq) + 2NH 3{ay) -» Ag{NH 3 ) 2(aq)

(b) H 2 {L) + NH 3{aq) -> NH 4 OH {aq)
Solution:

(a)Ag+ K) + 2NH 3[aq] -> Ag{NH z ) 2{aq)
Acid Base

H H H H

h : n : + Ag + : n : h — *■ h : n : Ag : n : h

H H H H

(a)tt 2 (L) + NH 3{aq) -> NH 4 OH {aq)
Acid Base

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H H

■ ■ ■■ ■ ■ ■■

h : n : + h:o: — > h : n :h:o:

■■ ■■ ■■ ■■

H H H H

Lesson Summary

• Lewis defined an acid as a substance that accepts a pair of eiectrons from a substance to form a
bond. Lewis denned a base as a substance that donates a pair of electrons to a substance to form a
bond.

• Coordinate covalent bond is the term given to bonds formed when both electrons come from the
same atom.

Review Questions

1. How do the Lewis definitions of acids and bases compare to the Br0nsted-Lowry definitions of acids
and bases?

2. In the following reversible reaction, which of the reactants is acting as a Lewis base? Cd 2+ i aq \ +

Ar {aq) u cdh 2 -

(aq)

(a)

Cd 2 +

(b)

(c)

Cdh 2 ~

(d)

none of the above,

this

not

acid-base reaction

3. Which of the following statements are false?

(a) NH 3 is a Lewis base.

(b) B{OH) 3 is a Lewis acid.

(d) Ag + Is a Lewis acid.

4. Which of the following statements are true?

(a) NH 3 is a Lewis base.

(b) B(OH) 3 is a Lewis acid.

(d) Ag + Is a Lewis acid.

5. Classify each of the following as a Lewis acid or base.

(a) H 2

(b) BF 3

(d) Cu 2+

(e) O 2 -

6. Write the balanced chemical equation between S0 3 2 ~ and H2O and label the Lewis acids and bases.

7. Identify the Lewis acid and Lewis base in each of the following reactions. Then write the reactions.

(a) Cu 2+ (aq) + 6 H 2 {L) -* Cu(H 2 0) 6 2 + {aq)

(b) (CH 3 CH 2 ) 2 (aq) +AlCl Haq) -» (CH 3 CH 2 )20AlCl 3{aq)

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Further Reading / Supplemental Links

• http://en.wikipedia.org

Vocabulary

Lewis acid A substance that accepts a pair of electrons from a substance (i.e. BF3).

Lewis base A substance that donates a pair of electrons from a substance (i.e. NH3).

coordinate covalent bond A covalent bond formed where both electrons that are being shared come
from the same atom.

Image Sources

(1) Examples of Lewis Acids..

(2) Theresa Forsythe. The pH Scale.. CC-BY-S A.

(3) pH Scale for Common Substances..

(4) Examples of Lewis Bases..

(5) Theresa Forsythe. . CC-BY-SA.

(6) Richard Parsons. Hydrated ions in solution.. CC-BY-SA.

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Chapter 18

Water, pH and Titration

18.1 Water Ionizes

Lesson Objectives

The student will write the equation for the autoionization of water and express the concentration of
hydrogen and hydroxide ion in a neutral solution at 25° C.

The student will express the value of K w in a water solution at 25°C.

The student will write the formulas for pH and pOH and show the relationship between these values
and K w .

The student will express the relationship that exists between pH,pOH, and K w .

Given the value of any one of the following values in a water solution at 25°C, the student will
calculate all the other values; [H + ], [OH + ], pH, and pOH.

The student will state the range of values for pH that indicate a water solution at 25°C is acidic.

The student will state the range of values for pH that indicate a water solution at 25°C is basic.

The student will state the range of values for pH that indicate a water solution at 25° C is neutral.

Introduction

There are many properties of water that we have learned. We know, for example, that water is called the
universal solvent because of its ability to dissolve both polar substances as well as most ionic ones. We also
know that pure water does not conduct electricity. The reason pure water does not conduct electricity is
because of the small concentration of ions present when water ionizes. In this lesson, we will look a little
closer at the uniqueness of autoionization. We will expand on the effect other added substances have on
the [//3<9 + ] and [OH~] in water. Thus, we need to have formulas readily available to calculate the pH, the
pOH, the [H^O + ], and/or the [OH~]. Much of what you will learn may be review.

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Water Ionization is Small But Significant

Autoionization, as we learned in a previous lesson, is the process where the same molecule acts as both
an acid and a base. Water is one of the compounds that can exhibit this unique property. When speaking
about the autoionization of water, another term that is also used is the self-ionization of water. The
reaction for the autoionization of water is shown in Equation 1 and its net ionic equation in Equation 2.

H 2 [L) + H 2 {L) £» H 3 + iaq) + OH- {aq) (Equation 1)

H + + OH + H 2 0±+ H 3 + (ag) + OH- (aq) (Equation 2)

If you notice in Equation 1, one H 2 Ou\ molecule donates a proton and is therefore a Br0nsted-Lowry acid;
the second H 2 On\ molecule accepts a proton and is therefore a Br0nsted-Lowry base.

While this does indeed occur, it only happens to about 2 molecules for every billion! This is the reason
that pure - or what can also be referred to as distilled or deionized water - will not conduct electricity. In
contrast, water that comes into our houses and we get from turning on the tap, is not pure water and can
conduct electricity. You may have seen hair dryers with warnings labels attached that remind you not to
use them while in the bath. Tap water has various ions dissolved in it, which makes it an electrolyte and
an electrical current that could pass through tap water and into you, could be dangerous. Some of the ions
that are dissolved in tap water are Na + , Cl~, Mg 2+ , Ca 2+ and various others. When tap water boils away
it often leaves a residue in the pan or kettle.

The Mathematics of pH and pOH

Water ionizes to a very slight degree.

H 2 {L) <=> H + + OH

At 25°C, the ionization equation above reaches equilibrium when [H + ] = [OH~] = 1 X 10 M. Therefore,
the equilibrium constant for this reaction, at 25°C, will be:

K w = [H + ][OH~] = (1 x 1(T 7 )(1 x 1(T 7 ) = 1 x 1(T 14

When the concentrations of the hydrogen and hydroxide ions are equal, we say the solution is neutral.
When some substance is added to the water and alters the hydrogen and hydroxide ion concentrations so
that they are no longer equal, the solution will become either acidic or basic. The hydrogen and hydroxide
ions will return to an equilibrium position but the concentrations of the ions may not be equal.

If, at equilibrium, [H + ] > [OH~], the solution is acidic.
If, at equilibrium, [H + ] = [OH~], the solution is neutral.
If, at equilibrium, [H + ] < [OH~], the solution is basic.

pH is used to express the acid strength of water solutions.

pH = -log [H + ]

For acid solutions, the pH will be less than 7. For basic solutions, the pH will be greater than 7. For
solutions that are neutral, the pH will be equal to 7.

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The pH equation can also be used to find the hydrogen ion concentration when the pH is given.

[H+] = l(T p//

Even though both acidic and basic solutions can be expressed by pH, someone decided to create an
equivalent set of expressions for the concentration of the hydroxide ion in water; hence pOH.

pOH = -log [0H~]

If the pOH is greater than 7, the solution is acidic. If the pOH is equal to 7, the solution is neutral. If the
pOH is less than 7, the solution is basic.

If we take the negative log of the complete K w expression:

K w =[H+][OH-}
-log K w = (-log [H+]) + (-log [OH-])
-log (1 x 1(T 14 ) = (-log [//+]) + (-log [OH-])
14 = pH + pOH.

We can determine that the sum of the pH and the pOH is always equal to 14 (at 25°C).

Remember that the pH scale is written with values from to 14 because many useful acid and base solutions
fall within this range. We can use this formula to carry many different types of calculations. Before moving
on to these, one more deduction can be made from all that we have done so far. If p() = — log() such that
pH = -\og[H + ] and pOH = —\og[OH~] and we know that K w = 1.00 x 10~ 14 , then we can calculate pK w
using the formula pK w = -log K w .

Now let's go through a few examples to see how this calculation works for problem-solving in solutions
other than water.

Sample question 1: What is the [H + ] for a solution of NH 3 whose [OH~] = 8.23 X 10~ 6 mol/L?

Solution:

[H 3 + ][OH-] = 1.00 xl0~ 14

.. 1.00 X10 -14 1.00 xlO -14 q

[H 3 + ] = = = = = = 1.26 x 10~ 9 M

L d J [OH-] [8.23X10- 6 ]

Sample question 2: Black coffee has a [H 3 + ] = 1.26 x 10~ 5 mol/L. What is the pOHl
Solution:

pH = -log [H + ] = -log 1.26 x 10~ 5 = 4.90
pH + pOH = 14

pOH =U-pH=U- 4.90 = 9.10

Lesson Summary

• Autoionization is the process where the same molecule acts as both an acid and a base:

• pK w = -logK w

. pH + pOH = pK w = 14.0

• pH + pOH = 14.0 is a formula used to find a number of different pieces of information from an acid-
base equation including [H + ], [OH~], pH, and pOH. Only one of these four pieces must be known to
find the other three.

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Review Questions

1. How do pH and pOH relate to the pH scale?

2. What does the value of K w tell you about the autoionization of water?

3. If the pH of an unknown solution is 4.25, what is the pOHl

(a) 1(T 4 - 25

(b) 1(T 9 - 75

(d) 14.0 -1(T 9 - 75

4. A solution contains a hydronium ion concentration of 3.36 X 10~4 mol/L. What is the pH of the
solution?

(a) 3.36

(b) 3.47

(d) none of the above

5. A solution contains a hydroxide ion concentration of 6.43 x 10~ 9 mol/L. What is the pH of the
solution?

(a) 5.80

(b) 6.48

(d) 8.19

6. An unknown solution was found in the lab. The pH of the solution was tested and found to be 3.98.
What is the concentration of hydroxide ion in this solution?

(a) 3.98 mol/L

(b) 0.67 mol/L

(d) 9.55 x 10' 11 mol/L

7. If a solution is known to have a hydroxide ion concentration of 2.5 X 10 -5 mol/L, then the pH of the
solution is and it is .

(a) 2.5, acidic

(b) 4.6, basic

(d) 9.4, basic

8. K w is the ionization product constant for water but is also the equilibrium constant for the acid-
base autoionization reaction for water. When dealing with equilibrium constants, such as K w , it is
important to take into account the temperature as temperature affects the value of the equilibrium
constant. The value of 1.0 x 10~ 14 for K w is for a temperature of 25° C. If the temperature was raised
to 60°C, the value of K w changes to 1.0 X 10~ 13 . How does this effect [H+],[OH~], pH, and pOHl

Further Reading / Supplemental Links

• http://en.wikipedia.org

Vocabulary

autoionization The process where the same molecule acts as both an acid and a base.
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18.2 Indicators

Lesson Objectives

• Define an acid-base indicator.

• Explain the difference between natural and synthetic indicators.

• List examples of natural and synthetic indicators.

• Explain how indicators work.

• Explain the usefulness of indicators in the lab.

Introduction

Indicators are used everyday both in our lives and in our laboratories. We make a cup of tea, we work in
the garden, we have a bowl of cereal and throw some blueberries on it, or we may go to chemistry class
and do an experiment to identify some of the properties of household substances using indicator solutions.
All of these situations involve the use of indicators; the only difference is that some indicators are natural
and some are synthetic (well, maybe - depending what you use in your chemistry class!) In this lesson you
are going to examine the concept of natural and synthetic indicators and the use of these in the modern
chemistry lab.

Substances That Change Color Due to pH Change

Litmus paper is a paper that has been dipped in a substance that will undergo a color change when it is
dipped in either an acid or a base. The litmus paper is called an indicator because it is used to indicate
whether the solution is an acid or a base. If the red litmus paper turns blue, the solution is basic {pH > 7),
if the blue litmus turns red the solution is acidic {pH < 7) .

Figure 18.1: Hydrangeas

An indicator is a substance that changes color at a specific pH and is used to indicate the pH of the
solution. The juice from red cabbage, for example, can be used to prepare an indicator paper. It contains
the chemical anthrocyanin which is the active ingredient in the indicator. Did you know that there are
actually indicator fish in the world? It's true! In Singapore there are fish that have been genetically altered
with the indicator in hopes of detecting pollutants in the water. These rainbow fish turn color from green
to red depending on the amounts of heavy metals and are thus given their name. (You may see, in news
articles, these fish referred to as "litmus fish" but, of course, the color changes in these fish have nothing to
do with the substance litmus.) Another example of a natural indicator is flowers. Hydrangea is a common
garden plant with flowers that come in many colors depending on the pH of the soil. If you are travelling

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around and see a hydrangea plant with blue flowers (Figure 18.1), the soil is acidic, the creamy white
flowers indicate the soil is neutral, and the pink flowers mean the soil is basic.

A natural indicator is an indicator that is produced from a substance that is naturally occurring or is
itself a naturally occurring substance. Red beets, blueberries, and cranberries are other great examples of
a naturally occurring indicator. These are all due to the same active ingredient anthocyanin found in the
red cabbage.

So how do naturally occurring indicators differ from synthetic indicators? Synthetic indicators are
compounds created in a chemistry lab rather than compounds found in nature. Both naturally occurring
indicators and synthetic indicators are weak organic acids or bases. For example, a common synthetic
indicator used in most chemistry laboratories is phenolphthalein (Figure 18.2).

Figure 18.2: Structure of Phenolphthalein.

This indicator, in particular, changes color at pH of 8.2, before 8.2 it is colorless and after 8.2 it is pink.
There are many common synthetic indicators that are useful in the chemistry laboratory. All of these
have similar structures. In the acidic range, chemistry students may use methyl orange. The structure for
methyl orange is shown in Figure 18.3.

O Na +

Figure 18.3: Structure of Methyl Orange (Yellow Form).

Methyl Orange changes color from pH 3.2 to 4.4. Below 3.2, the color of the indicator is red. Above 4.4
the color of the indicator is yellow. In between 3.2 and 4.4, at 3.8, the color would be orange ... hence the
name.

What happens chemically when there is a color change is that the structure of the molecule donates or
accepts a proton. Remember Br0nsted-Lowry acids and bases? A Br0nsted-Lowry acid donates a proton
and a Br0nsted-Lowry base accepts a proton. There are two requirements for a substance to function as
an acid-base indicator; 1) the substance must have an equilibrium affected by hydrogen ion concentration,
and 2) the two forms of the compound on opposite sides of the equilibrium must have different colors. Most

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indicators function in the same general manner and can be presented by a generic indicator equation. In
the equation below, we represent in the indicator ion with a hydrogen ion attached as HIn and we represent
the indicator ion without the hydrogen attached as In - .

The color of an indicator depends on the pH of the solution. (Source: Richard Parsons. CC-BY-SA)

ffln

fan)

«r

± H+ +

In"

Since the indicator itself is a weak acid, the equilibrium between the protonated form and the anionic form
is controlled by the hydrogen ion concentration. For the example above, the protonated form is colored
red and the anionic form is colored yellow. If we add hydrogen ion to the solution, the equilibrium will be
driven toward the reactants and the solution will turn red. If we add base to the solution (reduce hydrogen
ion concentration), the equilibrium will shift toward the products and the solution will turn yellow. It is
important to note that if this indicator changes color at pH = 5, then at all pH values less than 5, the
solution will be red and at all pH values greater than 5, the solution will be yellow. Therefore, putting this
indicator into a solution and having the solution turn yellow does NOT tell you the pH of the solution . . .
it only tells you that the pH is greater than 5 ... it could be 6, 7, 8, 9, etc. At pH values less than 5, the
great majority of the indicator molecules are in the red form and the solution will be red. At pH's greater
than 5, the great majority of the indicator particles will be in the yellow form and the solution will be
yellow. The equilibrium between these indicator particles is such that the particles will be 50% red form
and 50% yellow form at exactly pH = 5. Therefore, at pH = 5, the actual color of the solution will be a
50 - 50 mixture of red and yellow particles and the solution will be ORANGE.

The color change of an indicator occurs over a very short range. (Source: Richard Parsons. CC-BY-SA)

□ □ □

pH = 2

pH = 3 P H = 4

< < <

pH =

P H = 6

P H =

pH = 8

You should recognize that an indicator will only indicate the actual pH at one particular pH and that is
the color-change pH.

Laboratory Uses for pH Indicators

There are many indicators that are available to be used to help determine the pH of solutions. A list
of the most common indicators is found in Table 18.1 along with their respective color change pHs and
corresponding color changes.

Table 18.1: Colors and pH Ranges for Common Indicator

Indicator

pH range

Color Range

Methyl violet
Thymol blue
Orange IV
Methyl orange
Bromphenol Blue
Congo Red

0.0-1.6
1.2-2.8
1.3-3.0
3.2-4.4
3.0-4.7
3.0-5.0

Yellow - Blue
Red - Yellow
Red - Yellow
Red - Orange
Orange/Yellow
Blue - Red

Violet

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Table 18.1: (continued)

Indicator

pH range

Color Range

Bromocresol green
Methyl red
Litmus

Chlorophenol Red
Bromothymol blue
Phenol Red
Thymol blue
Phenolphthalein
Thymolphthalein
Alizarin yellow R
Methyl Blue
Indigo Carmine

3.8 - 5.4

4.8 - 6.0
5.0-8.0
4.8 - 6.2
6.0 - 7.6
6.4 - 8.2
8.0 - 9.6
8.2 - 10.0
9.4 - 10.6
10.1-12.0
10.6-13.4
11.4-13.0

Yellow - Blue
Red - Yellow
Red - Blue
Yellow - Red
Yellow - Blue
Yellow -Red/Violet
Yellow - Blue
Colorless - Pink
Colorless - Blue
Yellow - Red
Blue - Pale Violet
Blue - Yellow

There are many more indicators than are shown in the table above. Those that are shown are ones that
you may find in common chemistry classroom laboratories or in universities depending on where you are
located or what your experimental needs are at the time. One that is not found in the table is known as
the universal indicator. The universal indicator is a solution that has a different color for each pH from
0-14. Universal indicator is produced by creatively mixing many of the individual indicators together so
that a different coor is achieved for each different pH. It is used for all types of experiments to determine
if solutions are acids or bases and where on the pH scale the substance belongs. Figure 18.4 indicates the
colors of universal indicator for different pH values.

Color of Universal Indicator at Various pH Values

1 2

■

•

8 9

10 11

12 13 14

Figure 18.4: The color of Universal Indicator at various .

Why is this information be so important? If you look at Table 18.1, Thymol blue changes color from pH
of 8.0 to pH = 9.6. What this means is that from pH of to pH of 8.0, the color of the solution will stay
yellow. At pH = 8.0 the color will begin to change from yellow to blue and from pH of 9.6 until pH = 14,
the color of the solution will be blue. Around pH = 8.8 (the midway mark between 8.0 and 9.6) the color
of the solution would be green.

Sample question: If the pH of the solution is 4.8, what would be the color of the solution if the following
indicators were added?

(a) Universal indicator

(b) Bromocresol Green

(a) Universal indicator = Orange to orange-yellow (see Figure 18.4)

(b) Bromocresol Green = green (midway pH = 4.6)

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There is another important use for indicators in the lab and this is to determine the pH of an unknown
solution. They are also used in titrations, a concept we will be exploring in the another lesson. For now,
let's look at how we can use indicators to determine the pH of an unknown solution. For example, a
solution has been found in the laboratory that was tested with a number of indicators. It was found that
the following indicators showed these results:

Phenolphthalein was colorless

Bromocresol green was blue

Methyl red was yellow

Phenol red was yellow

What was the pH of the solution? Let's look at the data.

Phenolphthalein was colorless, pH < 8.0

Bromocresol green was blue, pH > 5.4

Methyl red was orange, pH > 6.2

Phenol red was yellow, pH < 6.4

Therefore the pH of the solution must be between 6.3 and 6.4.

Lesson Summary

• An indicator is a substance that changes color at a specific pH and is used to indicate the pH of
the solution. A natural indicator is an indicator that is produced from a substance that is naturally
occurring or is itself a naturally occurring substance.

• Synthetic indicators are normally an organic weak acid or base with usually a complicated structure.
Universal indicator is a solution that has a different color for each pH from to 14.

Review Questions

1. Describe the uses for litmus and universal indicator in the laboratory setting.

2. What is the difference between a natural and a synthetic indicator?

3. Describe how indicators work.

4. If you had an acid-base neutralization reaction that turned phenolphthalein pink and Thymolph-
thalein blue, what is the pH of the solution?

(a) 8.2

(b) 9.4

(d) Not enough information is available.

5. If you had an acid-base neutralization reaction that turned methyl violet blue and Thymol blue
orange, what is the pH of the solution?

(a) 1.6

(b) 2.0

(d) Not enough information is available.

6. Universal indicator is an indicator commonly used in the laboratory. At a pH of 6 it is pale yellow
and at a pH of 4 it is pale orange. If the indicator was orange, which statement would be definitely
true?

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(a) The solution is probably acidic.

(b) The pH is between 4 and 5.

(d) The pH is less than 5.0.

7. Alizarin Yellow 7? is an indicator that changes color in the pH range from 10.1 to 12.0. Below 10.1
the color is Yellow, above 12.0 the color is red. If the color of the solution containing Alizarin Yellow
R was orange, which statement about the solution would be true?

(a) The pH is below 10.

(b) The pH is above 12.0.

(d) The pH is between 10.1 and 12.0.

8. If the pH of the solution is 8.9, what would be the color of the solution if the following indicators
were added?

(a) Universal indicator

(b) Thymol blue

9. A solution has been found in the laboratory that was tested with a number of indicators. It was
found that the following indicators showed these results:

(a) Phenolphthalein was colorless

(b) Orange IV was yellow

(d) Methyl orange was red

What was the pH of the solution?

Further Reading / Supplemental Links

• http://en.wikipedia.org

Vocabulary

indicator A substance that changes color at a specific pH and is used to indicate the pH of the solution.

natural indicator An indicator that is produced from a substance that is naturally occurring or is itself
a naturally occurring substance.

synthetic indicator An indicator that is a complicated structure of an organic weak acid or base.

Possible Laboratory Activities for Indicators
Teacher's Pages for pH Measurements Using Indicators

Investigation and Experimentation Objectives

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In this activity, the student will observe the relationships between theoretical mathematical conclusions
and laboratory results.

Lab Notes

Buffered solutions of various pH values can be purchased in dropper bottles, as can dropper bottles of
indicator solutions. If you have many chemistry classes and perform the experiment for several years, it
may be more economical to prepare the solutions yourself.

The solutions used in this lab can be prepared as follows.

pH = 1 solution: dilute 8.3 mL of concentrated HCl (12 M) to 1.00 liter
pH = 3 solution: dilute 10. mL of pH = 1 solution (above) to 1.00 liter
pH = 5 solution: dilute 10. mL of pH = 3 solution (above) to 1.00 liter
pH = 7 solution: distilled water

pH = 13 solution: dissolve 4.00 g of NaOH in sufficient water to produce 1.00 liter of solution
pH = 11 solution: dilute 10. mL of pH = 13 solution (above) to 1.00 liter
pH = 9 solution: dilute 10. mL of pH = 11 solution (above) to 1.00 liter

methyl orange indicator: dissolve 0.1 g of methyl orange powder in 100 mL of water and filter
bromthymol blue indicator: dissolve .01 g of bromthymol blue in 100 mL of 50% water and 50%
ethanol solution and filter
• phenolphthalein solution: dissolve 1.0 g of phenolphthalein powder in 100 mL of ethanol

To prepare the unknown solutions for Part IV, select three of the known pH solutions used in the lab and
label them as unknown. Be sure to keep a record of which pH values were selected as unknowns.

Answers to Pre-Lab Questions

1. How is universal indicator made?
Several indicators are mixed.

2. What distinguishes weak organic acids that are useful as acid-base indicators from weak organic acids
that will not function as acid-base indicators?

The undissociated molecule of the acid and the anion of the dissociated acid must be different colors.

pH Measurements Using Indicators

Background Information

The nature of acids and bases have been known to man for quite sometime. Chemically speaking, acids
are interesting compounds because a large number of common household substances are acids or acidic
solutions. For example, vinegar contains ethanoic acid, also called acetic acid, //C2//3O2) and citrus fruit
contain citric acid. Acids cause foods to have a sour taste and turn litmus red. (Note: You should never
taste substances in the laboratory.) Also, many common household substances are bases. Milk of magnesia
contains the base magnesium hydroxide, Mg{OH)2 and household ammonia is a common cleaning agent.
Bases have a slick feel to the fingers and turn litmus blue. (Note: You should never feel chemicals in the
laboratory.)

Indicator dyes, of which litmus is one, turn various colors according to the strength of the acid or base
applied to it.

Pure water, which is neutral in terms of acid-base, exists mostly as H2O molecules but does, to a very
slight extent dissociate into hydrogen and hydroxide ions.

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The extent of this dissociation is 1.0x10 7 moles /liter (at 25°C). Therefore, in all neutral water (and neutral
water solutions), the concentration of hydrogen ions is 1.0xl0~ 7 M and the concentration of hydroxide ions
is 1.0 x 10~ 7 M. The dissociation constant for this process is K w = [H + ][OH~] = (1.0 x 10~ 7 )(1.0 x 10~ 7 =
1.0 xlO" 14 ).

In 1909, a Danish chemist (Soren Sorenson), developed a mathematical system for referring to the degree
of acidity of a solution. He used the term pH for "power of hydrogen" and established the equation,
P H = -log[H+}.

In a neutral solution, the hydrogen ion concentration is 1.0 X 10~ 7 M and therefore, the pH is 7. If the
concentration of hydrogen ions is 1.0 X 10 -5 M, then the pH is 5. A solution is neutral when the pH equals
7, it is acid if the pH is less than 7, and it is basic if the pH is more than 7. In commonly used solutions,
pH values usually range from 1 to 14.

Living matter (protoplasm) contains a mixture of variously dissociated acids, bases, and salts and usually
has a pH very near neutral. The pH of human blood is generally 7.3 and humans cannot survive if the
blood becomes more basic than pH 7.8 or more acidic than pH 7.0. Life of any kind exists only between
pH 3 and pH 8.5. Buffer solutions regulate the pH of the body by neutralizing excess acid or base. The
chief buffers of the body are proteins, carbonates, phophates, and hemoglobin. The kidneys play a role by
eliminating excess electrolytes.

Table 18.2: Some Typical

Substance

Acidity /Basicity

1
2

Sulfuric Acid (Battery Acid)
0.10 M Hydrochloric Acid
Stomach Acid

Vinegar

Tomato Juice

Black Coffee and Vitamin C

Cow's Milk

Distilled Water

Sea Water

Baking Soda, NaHCOj,

Detergents

Household Cleaning Ammonia

0.10 M NaOH

1.0 M NaOH (Lye)

Very Highly Acidic
Highly Acidic
Acidic

Acidic (jj^'

as strong as pH 1]

Weakly Acidic

Very Weakly Acidic

Neutral

Very Weakly Basic

Weakly Acidic

Basic

Strongly Basic
Very Strongly Basic

Some acids dissociate completely into ions when dissolved in water. Such acids are called strong acids
(HCl, HI, HBr, HNO%, H2SO4, HCIO4). Some bases dissociate completely when dissolved in water. Such
bases are called strong bases (NaOH, LiOH, KOH, RbOH). There are other acids and bases that dissociate
only slightly (although completely soluble) when dissolved in water. Such acids and bases are called weak
acids or weak bases and some examples are HF , HC2H3O2, NH4OH.

An important method of determining pH values in the lab involves the use of substances called "acid-base
indicators". These are certain organic substances (almost always weak organic acids) that have the property
of changing color in solutions of varying hydrogen ion concentration. In order for a weak organic acid to
be useful as an acid-base indicator, it is necessary that the undissociated molecule and the indicator anion
be different colors. For example, phenolphthalein is a colorless substance in any aqueous solution in which
the hydrogen ion concentration is greater than 1 X 10 -9 M (pH < 9) but changes to a red or pink color

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when the hydrogen ion concentration is less than 1 x 10 -9 M (pH > 9). Such substances can be used for
determining the approximate pH of solutions. Electrical measurements can determine the pH even more
precisely. This lab will use three acid base indicators and what is called a "universal indicator".

Table 18.3: Some Indicator Color Changes

Indicator pH Color Change Range Color Change

Methyl Orange 3.1 - 3.4 Red to Yellow

Bromthymol Blue 6.0 - 7.6 Yellow to Blue

Phenolphthalein 8.3 - 10.0 Colorless to Red

The universal indicator (one type is called Bogen's Universal Indicator), is made by mixing a number of
indicators that all change color at different pH's. As you slowly change the pH of the indicator from 1
to 14, it goes through a series of subtle color changes. The indicator is provided with a photographic
chart that shows the color of the indicator at every different pH and the pH is identified by matching the
indicator color to the chart.

Pre-Lab Questions

1. How is universal indicator made?

2. What distinguishes weak organic acids that are useful as acid-base indicators from weak organic acids
that will not function as acid-base indicators?

Purpose

The purpose of this lab is to have the student experience the color changes involved with acid-base indicators
and to identify the approximate pH of an unknown solution using acid-base indicators.

Apparatus and Materials

Well Plates, at least 12 wells (1 per lab group)

drop controlled bottles of pH = 1

drop controlled bottles of pH = 3

drop controlled bottles of pH = 5

drop controlled bottles of pH = 7

drop controlled bottles of pH = 9

drop controlled bottles of pH = 11

drop controlled bottles of pH = 13

drop controlled bottles of methyl orange

drop controlled bottles of bromthymol blue

drop controlled bottles of phenolphthalein

drop controlled bottles of universal indicator

drop controlled bottle of unknown #1

drop controlled bottle of unknown #2

drop controlled bottle of unknown #3

Safety Issues

All solutions are irritating to skin, eyes, and mucous membranes. Handle solutions with care, avoid getting
the material on you, and wash your hands carefully before leaving the lab.

Procedure for Part I: Determining the effect of pH on indicator dyes.

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1. Place the Chemplate on a sheet of white paper.

2. Place one drop of methyl orange into cavities #1 and #2.

3. Place one drop of bromthymol blue in cavities #5 and #6.

4. Place one drop of phenolphthalein in cavities #9 and #10.

5. Carefully add one drop of pH 1 to cavities #1, #5, and #9.

6. Carefully add one drop of pH 13 to cavities #2, #6, and #10.

Data for Part I

1. What color is the original methyl orange solution?

2. What color is methyl orange in a strong acid?

3. What color is methyl orange in a strong base?

4. What color is the original bromthymol blue solution?

5. What color is bromthymol blue in a strong acid?

6. What color is bromthymol blue in a strong base?

7. What color is the original phenolphthalein solution?

8. What color is phenolphthalein in a strong acid?

9. What color is phenolphthalein in a strong base?

Rinse the Chemplate in tap water and dry with a paper towel.

Procedure for Part II: Determining the pH color change range of indicator dyes.

1. Place one drop of methyl orange in each cavity numbered 1-7.

2. Carefully add one drop of pH 1 to cavity #1, pH 3 to cavity #2, pH 5 to cavity #3, pH 7 to #4,
pH 9 to #5, pH 11 to #6, and pH 13 to #7.

3. Repeat the rinsing, drying, and steps 1 and 2 except using bromthymol blue and then repeat the
entire process again using phenolophthalein.

Data for Part II

1. Describe the color changes and the pH 's around the color change pH for methyl orange.

2. Describe the color changes and the pH 's around the color change pH for bromthymol blue.

3. Describe the color changes and the pWs around the color change pH for phenolphthalein.

Rinse the Chemplate in tap water and dry with a paper towel.

Procedure for Part III: Determining a color standard for universal indicator.

1. Place one drop of universal indicator in each cavity numbered 1-7.

2. Carefully add one drop of pH 1 to cavity #1, pH 3 to cavity #2, pH 5 to cavity #3, pH 7 to #4,
pH 9 to #5, pH 11 to #6, and pH 13 to #7.

Keep these solutions for Part IV.
Data for Part III

1. Describe the color of the universal indicator at each pH used.

Cavity #1 (pH — 1), color =

Cavity #2 (pH = 3), color =

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Cavity #3 (pH = 5), color =
Cavity #4 (pH = 7), color =
Cavity #5 (pH = 9), color =
Cavity #6 (pH =11), color
Cavity #7 (pH = 13), color

Procedure for Part IV: Determining the pH of some unknown solutions.

1. Place one drop of universal indicator in cavities #10, #11, and #12.

2. Place one drop of unknown #1 in cavity #10.

3. Place one drop of unknown #2 in cavity #11.

4. Place one drop of unknown #2 in cavity #12.

5. Compare the color in each cavity with the colors in cavities #1-7 that you made in activity 3.

Data for Part IV

Color of unknown #1 in universal indicator

Color of unknown #2 in universal indicator

Color of unknown #3 in universal indicator

Post-Lab Questions

1. What is the pH of unknown #1?

2. What is the pH of unknown #2?

3. What is the pH of unknown #3?

18.3 Titrations

Lesson Objectives

Define titrations and identify the different parts of the titration process.

Explain the difference between the endpoint and the equivalence point.

Describe the three types of titration curves.

Identify points on the titration curves for the three types of titrations.

Define a standard solution.

Calculate the accurate concentration of an acid or base using a standard.

Calculate unknown concentrations or volumes of acids or bases at equivalence.

Introduction

For acid-base neutralization reactions, the typical laboratory procedure for determining the stoichiometric
amounts of acid and/or base in the reaction is to complete a titration. There are three main types of
titration experiments and each of these has some similarities but also have some differences. A titration
experiment can involve equipment from the simple (such as using an eyedropper) to the complex (such as
using a burette, a pH meter, a magnetic stirrer). As we go through this lesson, we will use some of the
prior knowledge we have obtained about acids and bases, about chemical reactions, molarity calculations,
and about indicators to apply them to the concept of titrations.

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Titration

One of the properties of acids and bases is that they neutralize each other. In the laboratory setting,
an experimental procedure where an acid is neutralized by a base (or vice versa) is known as a titration.
A titration, by definition, is the addition of a known concentration of base (or acid) (also called the
titrant) to a solution of acid (or base) of unknown concentration. Since both volumes of the acid and base
are known, the concentration of the unknown solution is then mathematically determined. So what does
one do in a titration? When doing a titration, you need to have a few pieces of equipment. A burette
(Figure ??) is used to accurately dispense the volume of the solution of known concentration (either the
base or the acid). An Erlenmeyer flask is used to hold a known volume of the unknown concentration
of the other solution (either the acid or the base). Also an indicator is used to determine the endpoint
of the titration. A few drops of the indicator are added to the flask before you begin the titration. The
endpoint is the point where the indicator changes color, which tells us that the acid is neutralized by the
base. The equivalence point is the point where the number of moles of acid exactly equals the number
of moles of base. The equivalence point is a calculated point in the neutralization of the acid and the base.
Some laboratories have pH meters (Figures 18.5, 18.7) that measures this point more accurately than
the indicator although an indicator is much more visual! Figure 18.5 shows a simplified version of a pH
meter with the probe from the meter immersed in the liquid. Figure 18.6 shows a typical electronic pH
meter with the attached probes. The main purpose of a pH meter is to measure the changes in pH as the
titration goes from start to finish.

Burette. (Source: http://commons.wikimedia.Org/wiki/File:Burette.png. Public Domain)

Shown above is a typical titration setup. The burette is upright ready to drip the solution into the flask
holding the solution of unknown concentration and the few drops of indicator. When the indicator changes
color, the number of moles of acid equals the number of moles of base and the acid (or base) has been
neutralized.

There are three types of titrations that are normally performed in the laboratory in order to determine
the unknown concentration of the acid or base. These three types are:

1. Strong acid vs. Strong base
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•

8.03

• •

Figure 18.5: A simple meter with its probe immersed in a mildly alkaline solution . The two knobs on the
meter are used to calibrate the instrument.

Figure 18.6: An electronic meter

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■_
:

:
7

Figure 18.7: Typical Titration Setup.

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2. Strong acid vs. Weak base

3. Weak Acid vs. Strong base

In these titrations, a pH meter may be used to measure the changes in the pH as the titration goes to
completion. If so, a titration curve can be constructed for each curve. A titration curve is a graph of the
pH versus the volume of titrant added. Let's take a look at how each of these types of titrations differs in
terms of their pH curves and their pH at the equivalence point.

(1) Strong Acid vs. Strong Base

For a strong acid vs. a strong base titration, let's assume the strong base is the titrant. Therefore, in
the Erlenmeyer flask is the strong acid and a few drops of your indicator. Think about the pH of the
solution in the flask. Do you think it will be high (around pH = 11.0), mildly basic (around pH = 8.0),
mildly acidic (around pH = 6.0), or low (around pH = 1.0)? Probably the pH will be around 1.0 since the
solution is a strong acid. As the base is added, the acid is slowly neutralized. At first the change in pH is
minimal. This resistance is due to the fact that the flask has a much greater number of H^O + ions than
the OH~ ions available from the added titrant.

As more and more base is added, more OH ions are added and thus more H%0 + ions get neutralized.
Let's stop here and look at the reaction. Equation 1 shows the total ionic equation of a reaction between
a strong acid and a strong base.

H+ (aq) + CI (aq) + Na+ (aq) + OH ( ag ) -» NcT \ aq ) + CI ( aq ) + H 2 0( L )

(Equation 1)

Equation 2 shows the net ionic equation for the reaction between the strong acid and the strong base.
H + {aq ) + OH- {aq) -> H 2 {L) (Equation 2)

So you see that as we add more OH~ ions, more H^O + (or H + ) ions are being neutralized. Since these two
ions react to form water, a neutral solution will be formed. For a strong acid and a strong base, this means
the pH = 7.0 at the point of neutralization. If we continue to add the titrant (containing O// - ions) after
all of the H^O + ions have been neutralized, the pH will continue to rise as more base is added and there
are excess OH~ ions.

Now we know what happens in a strong acid/strong base titration, what does the titration curve look like?
Look at Figure 18.8. The main points of the titration curve described above are shown in the titration
curve below.

D Equivalence point

Volume of titrant added

Figure 18.8: Titration Curve for a Strong Acid vs. Strong Base.

The points A through D sum up the description of the events that take place during the titration. Point
A is the start of the titration. Point B is the midpoint, the point where half of the H + ions have been
neutralized. Point D is the equivalence point.

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(2) Strong Acid vs. Weak Base

What would happen if we were to titrate a strong acid with a weak base or vice versa? Look at Figure
18.9. Can you determine what is happening in the titration just by looking at the graph?

Equivalence point

Volume of titrant added

Figure 18.9: Titration Curve for a Strong Acid vs. Weak Base.

When you look at Point A, the pH is high but not too high {pH approx 11). This means that there is a
weak base in the flask. As the acid (the titrant) is added, the pH decreases as the H 3 + ions being added
and begin to neutralize the OH~ ions. Point C represents the point where the volume of acid necessary to
neutralize the base is measured. Therefore, if we draw a line down to the x-axis, we can find the volume of
titrant necessary for the neutralization reaction (same as before). Point D is the equivalence point. Notice
that for a weak base and a strong acid titration, the pH at equivalence point is more acidic. Equation 3 is
the reaction between NH

3(aq)i a weak base, and HClr aq \, a strong acid.

NH 3(aq) + HCl(aq) "» ^ H 4 C l(aq) + #20 (L)

The ionic equation is Equation 4.

NH 3(aq) + h+ cj) + cr (uq) -> NH 4 + {aq) + cr {aq) + H 2 {L)

(Equation 3)

(Equation 4)

(3) Weak Acid vs. Strong Base

The third type of titration is that of a weak acid with a strong base. When we follow through with the
same procedure as was done with the previous two titrations, we can determine a great deal of information
simply by looking at the pH curve. For example, let's consider the titration of a solution of acetic acid,
HC2H3O2, with a solution of potassium hydroxide, KOH. We can write the chemical reaction for this
acid-base neutralization (see Equation 5) and begin to draw a rough sketch of a titration curve.

H+ (aq) + C 2 H 3 2( + K + {aq) + OH {aq) -> K + {aq) + C 2 H 3 2( + H 2 {L)

(Equation 5)

Acetic acid is a weak acid and we know that the pH at the start of the titration will be around 2.8.
Potassium hydroxide is a strong base so the curve will end high on the graph, around a pH = 12.

p. Equivalence point

Volume of titrant added

The points on the curve represent the same points as with the other two titration curves. Look, however, at
the equivalence point. Notice how the pH for the equivalence point of the weak acid- strong base titration
is above 7.0. Look at Equation 5.

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Sample question: Draw a rough sketch of the titration curve between nitric acid and ethylamine, CH3NH2.
Assume the acid is in the burette. What is the estimated pH at the equivalence point?

Solution:

Volume of titrant added

The pH at the equivalence point is approximately 4.6 from this graph.

Standard Solutions

As was previously stated, when we do a titration the titrant is the solution of known concentration. For
accuracy reasons, this titrant is normally titrated to find the exact concentration before beginning the
acid-base titration. The purpose of this initial titration is to determine, with as much accuracy as possible,
the exact concentration of the solution in the burette. To determine the exact concentration of the titrant,
we use a standard solution. A standard solution is a solution whose concentration is known exactly.
Standard solutions have this property because these chemicals are normally found in pure, stable forms.
Examples of chemicals used to prepare standard solutions are the acidic potassium hydrogen phthalate,
KHC 7 H^O^ (sometimes referred to as KHP), and the basic sodium carbonate, NCI2CO3.

When using a standard solution, the standard is first prepared by dissolving the solid in a known volume
of water, add a few drops of indicator, and titrate with the solution that you want to standardize.

Sample question: What is the concentration of sodium hydroxide when 32.34 mL is required to neutralize
a solution prepared by dissolving 1.12 g of KHC 7 H/^0^ in 25.00 mL of H^Ofj^l

Solution:

Step 1: Find the mols of KHC 7 H^O^t s y

mass 1.12 g ,

mols KHC 7 H 4 4 = — = f- — - = 5.83 x 10~ 3 mol

molar mass 192.2 g/mol

Step 2: Use mol ratio from the reaction to find the moles of NaOH.

KHC 7 H±0 A{aq) + NaOH {aq) -> KNaC 7 H 4 4{aq) + H 2 {L)

Since the reaction is 1:1, or 1 mole of KHP reacts with every mole of NaOH, the number of moles of
KHP = number of moles of NaOH.

mol NaOH = 5.83 X 10~ 3 mol

Step 3: Determine the concentration of NaOH.

. 5.83 x 10~ 3 mol

NaOH] = = 0.180 M

L J 0.03234 L

Therefore, the exact concentration of the sodium hydroxide solution used in the titration is 0.180 mol/L.

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Choosing an Appropriate Indicator

To choose an appropriate indicator for a titration, a titration curve is useful. Knowing the pH at equivalence
for the different types of titrations (see Table 18.4) is also needed.

Table 18.4: pH at Equivalence for Titrations

Type of Titration

pH at Equivalence

Strong Acid - Strong Base
Strong Acid - Weak Base
Weak Acid - Strong Base

pH = 7.0
pH < 7.0
pH > 7.0

Choosing an indicator close to the equivalence point is essential to see the point where all of the H + ions
and OH" ions have been neutralized. The color change should occur on or around the equivalence point.
So, for example, with a strong acid, strong base titration, the pH at equivalence is 7.0. Indicators such
as broniothymol blue [pH range = 6.0 - 7.6) and phenol red {pH range = 6.6 - 8.0) are common. If you
notice the midpoint color (green) for broniothymol blue would appear at a pH = 6.8 which is close to 7.0.
For phenol red, the midpoint color (orange) would appear at pH = 7.3, again close to 7.0.

The same process is used for other titration types. For a strong acid-weak base titration where the pH at
equivalence is less than 7, the indicators normally chosen for these titrations are methyl red {pH range =
4.8 — 6.0) and chlorophenol red {pH range = 4.8 - 6.2). For a weak acid-strong base titration where the pH
at equivalence is greater than 7, the indicators normally chosen for these titrations are phenolphthalein
{pH range = 8.2 - 10) and thymol blue {pH range = 8.0 - 9.6). As with strong acid-strong base titration,
the visual observation of the midpoint color should indicate close proximity to the equivalence point.

Sample question: Look at the graph below and determine the appropriate indicator.

Graph 1: Titration of a Weak Acid with a Strong Base (Source: Therese Forsythe. CC-BY-SA)

Solution:

We first look at the graph and mark the vertical stretch of the titration curve (red lines) in order to find
the half-way mark on this vertical stretch (green line). Looking at the graph, when we follow this half-way
mark over to the y-axis, we can see that the equivalence point occurs at approximately pH = 8.2. The
indicator appropriate to use would be phenolphthalein {pH range = 8.2 - 10) and as soon as the pink color
forms we are at the equivalence point.

Graph 2: Titration of a Weak Acid with a Strong Base (Source: Therese Forsythe. CC-BY-SA)

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Reaction Continues to the End Point

There is an interesting observation about the endpoint that has yet to be mentioned. The endpoint was
defined earlier as the point where the indicator changes color. In an acid-base neutralization reaction, this
point may not be the point where all of the H + ions have been neutralized by OH" ions, or vice versa. The
experimenter continues titration until the indicator changes color, that is, the endpoint has been reached.
The equivalence point is the point where the moles of hydrogen ion and the moles of hydroxide ion are
equal. It requires knowledge by the experimenter to select an indicator that will make the endpoint as
close as possible to the equivalent point.

Titration Calculations

For the calculations involved here, we will restrict our acid and base examples where the stoichiometric
ratio of H + and OH~ is 1 : 1. To determine the concentration or volume required to neutralize and acid or
a base, in other words, to reach the equivalence point, we will use a formula similar to the dilution formula
used in your prior learning. The formula has the structure:

M a x V a = M b x V h

— where —

M a = molarity of the acid

V a = volume of the acid
Mf, = molarity of the base

Vb = volume of the base

Sample question: When 10.0 mL of a 0.125 mol/L solution of hydrochloric acid, HCl, is titrated with a
0.100 mol/L solution of potassium hydroxide, KOH, what is the volume of the hydroxide solution required
to neutralize the acid? What type of titration is this?

Solution:

Step 1: Write the balanced ionic chemical equation.

H (aq) + CI ( m/ ) + K ( ag ) + OH ( ag ) -> K ( a? ) + CI ( a? ) + H 2 0( L -)

Step 2: Use the formula and fill in all of the given information.

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M a x V a = M h x V h

M a = 0.125 mol/L

V a = 10.0 mL

M h = 0.100 mol/L

V* = ?

M a xV a = M b x V fo

M fl xV fl (0.125 mol/L) (10.0 mL)

K/, = = ; = 12.5 mL

M b 0.100 mol/L

Therefore, for this weak acid-strong base titration, the volume of base required for the titration is 12.5 mL.

Lesson Summary

• A titration is the addition of a known concentration of base (or acid) to a solution of acid (or
base) of unknown concentration. The titrant in the titration is the solution of known concentration.
This solution is normally in the burette. A burette is a piece of equipment used in a titration to
accurately dispense the volume of the solution of known concentration (either the base or the acid) .
The endpoint is the point in the titration where the indicator changes color. The equivalence point
is the point in the titration where the number of moles of acid equals the number of moles of base;
pH meters are used in many laboratories for acid-base titrations. The main purpose of a pH meter
is to measure the changes in pH as the titration goes from start to finish.

• The three types of titrations usually performed in the laboratory are: strong acid vs. strong base,
strong acid vs. weak base, and weak acid vs. strong base. A titration curve is a graph of the pH
versus the volume of titrant added. For a strong acid vs. strong base titration, the pH at equivalence
is 7.0. For a strong acid vs. weak base titration, the pH at equivalence is less than 7.0. For a
weak acid vs. strong base titration, the pH at equivalence is greater than 7.0. A standard solution
is a solution whose concentration is known exactly and is used to find the exact concentration of
the titrant. For titrations where the stoichiometric ratio of mol H + : mol OH~ is 1 : 1, the formula
M a x V a = Mb x Vb can be used to calculate concentrations or volumes for the unknown acid or base.

Review Questions

1. Why do you think there would be more experimental error with using an indicator over using a pH
meter in a titration?

2. Why would there not be a weak acid- weak base titration?

3. Which of the following definitions best suits that of an endpoint?

(a) The stoichiometric point where the number of moles of acid equals the number of moles of base.

(b) The visual stoichiometric point where the number of moles of acid equals the number of moles
of base.

(d) None of the above

4. In the following titration curve, what pair of aqueous solutions would best represent what is shown
to be happening in the curve?

(a) HCOOH (aq) + NH 3{aq)
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(b) HCOOH (aq) + NaOH (aq)

I A\ Ufln . , . 1 Mil- , .

(d) HClO A{aq) + Wtt;

3 (a?)

1
80

i i r

20 40

Volume(mL)

5. What would be the best indicator to choose for the pH curve shown in question 3?

(a) Methyl red

(b) Litmus

(d) Phenol red

6. What is the best indicator to use in the titration of benzoic acid with barium hydroxide?

(a) Methyl violet

(b) Bromothymol blue

(d) Methyl blue

(e) Indigo carmine

Table 18.5:

Indicator

pH Range

Methyl violet
Bromothymol blue
Phenolphthalein
Methyl blue
Indigo carmine

0.0-1.6
3.0-4.7
8.2 - 10.0
10.6-13.4
11.4-13.0

7. If 22.50 mL of a sodium hydroxide is necessary to neutralize 18.50 niL of a 0.1430 mol/L HNO3
solution, what is the concentration of NaOHl

(a) 0.1176 mol/L

(b) 0.1430 mol/L

(d) 2.64 mol/L

8. Plot the following titration data on a titration curve of pH vs. volume of base added. When complete,
find the pH at equivalence and choose an appropriate indicator for the titration. What volume of

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base is necessary to neutralize all of the acid?

Table 18.6:

Volume of base added (mL) pH

0.00 1.0

2.00 1.2

4.00 1.4

6.00 1.6

8.00 1.9

9.00 2.3

9.50 2.6

9.90 3.3

9.99 4.3

10.00 7.0

10.01 9.7
10.50 10.7
12.00 11.4
14.00 12.1
16.00 12.1

9. Calculate the concentration of hypochlorous acid if 25.00 mL of HCIO is used in a titration with
32.34 mL of a 0.1320 mol/L solution of sodium hydroxide.

Further Reading / Supplemental Links

http : //en . wikipedia . org

Vocabulary

titration The lab process in which a known concentration of base (or acid) is added to a solution of acid
(or base) of unknown concentration.

titrant The solution in the titration of known concentration.

burette A piece of equipment used in titrations to accurately dispense the volume of the solution of
known concentration (either a base or an acid).

Erlenmeyer flask A piece of equipment used in titrations (and other experiments) to hold a known
volume of the unknown concentration of the other solution (either the acid or the base).

endpoint The point in the titration where the indicator changes color.

equivalence point The point in the titration where the number of moles of acid equals the number of
moles of base.

pH meter A device used to measure the changes in pH as the titration goes from start to finish.

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titration curve A graph of the pH versus the volume of titrant added.

standard solution A solution whose concentration is known exactly and is used to find the exact con-
centration of the titrant.

Labs and Demonstrations for Titrations
Teacher's Pages for Acid-Base Titration

Investigation and Experimentation Objectives

In this activity, the student will use burets, pipets, and acid-base indicators to collect data. The properly
recorded data with then be used along with several mathematical formulas to determine unknown molecular
masses from titration data.

Lab Notes

Students will need two days to do both parts of the lab.

Preparation of Solutions, KHP, and unknown acids

6 M NaOH solution: Boil 600 mL of distilled water to drive off any dissolved COi- (The CO2 produces
carbonic acid, which drives down the concentration of NaOH.) Add 120. g of NaOH to a 500 mL volumetric
flask, and add some of the freshly boiled distilled H20 to the flask. Swirl to dissolve. Cool the resultant
solution in a cold water or ice water bath, let the solution and flask return to room temperature, and dilute
the resulting solution to 500 mL. Store this solution in a tightly capped bottle, preferably Nalgene or other
base-resistant bottle. This will provide enough solution for (75) two-student teams with a 20% excess for
spills and endpoint over-runs.

Phenolphthalein indicator solution: Dissolve 0.1 g of phenolphthalein in 50 mL of 95% ethanol, and dilute
to 100 mL by adding distilled water. This will provide enough solution for (75) two student teams.

KHP - Potassium Hydrogen Phthalate - Dry 200 g of KHP in a laboratory oven for at least one hour prior
to titration. Store the KHP in a dessicator. KHP is slightly hygroscopic. This is sufficient KHP for (75)
two - student teams.

Unknown Acids

The following acids are suggestions for use. Their number of ionizable hydrogens vary, are stable chemically,
and many schools have them on hand. Twenty grams of each acid are required for (75) two - student teams:

Table 18.7: The Number of Ionizable Hydrogens and Molar Mass of Various Acids

Acid Number of Ionizable Hydro- Molar Mass (g/mol)

gens

Lactic 1 90.1

Malonic 2 104.1

Maleic 2 116.1

Succinic 2 118.1

Benzoic 1 122.1

Salicylic 1 138.1

Tartaric 2 150.1

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NOTE: The molecular weights listed are for the anhydrous acids. If hydrates are used, use the molecular
weights as shown on the reagent bottle. The number of ionizable hydrogens does not change.

Answers to Pre-Lab Questions

1. Pre-rinsing the buret will remove any water or residual NaOH solution within the buret. If there were
water present, it would dilute the NaOH added. If there were residual NaOH, the concentration would
increase upon addition of solution due to the crystalline or concentrated solution of NaOH present.

2. The KHP and the NaOH react with each other in a 1:1 stoichiometric ratio. Thus, the number of moles
of NaOH required to react with the KHP will be equal to the number of moles of KHP originally present.

Moles of NaOH added = M x L = (0.1 M)(0.040 L) = 0.004 moles NaOH added: thus

Moles KHP needed = 0.004 moles
Grams of KHP needed = (mole s)(g/ mole) = (0.004 mot) (204.23 g/mol) = 0.8 grams KHP

3. Molarity of KHP - moUs KHP

liters of solution

Moles of KHP = JZZfKHP = 2oI?fc = °-°° 372 m ° l KHP

Molarity = 0003 7 g 5 ^ / / //p = 0.0743 M KHP

4. Again, the amount of water needed to dissolve the KHP is not needed to solve this problem. All you
need is the weight of the KHP and it's molecular weight :

Moles of KHP = ' i— = 0.00255 mols KHP

J 204.2 g/mol

Since this will be equal to the number of moles of NaOH reacted, the number of liters of titrant you need
to add can be calculated by re-arranging the molarity formula:

moles

M =

moles 0.00255 mol

so L = = — = 0.0213 L = 21.3 mL

M 0.102 mol/L

Acid-Base Titration Lab

Background Information

Standardization of the sodium hydroxide solution through titration is necessary because it is not possible
to directly prepare a known molarity solution of sodium hydroxide with high accuracy. Solid sodium
hydroxide readily absorbs moisture and carbon dioxide from the atmosphere and thus it is difficult to
obtain a precise amount of the pure substance. A sodium hydroxide solution will be made close to 0.1 M
and then the actual molarity of the solution will be determined by titration of a primary standard. A
primary standard is a substance of very high purity that is also stable in air. Because the substance
remains pure, it is possible to mass a sample of the substance with a high degree of accuracy. The primary
standard used in this experimental procedure is potassium hydrogen phthalate, KHCgH^O^ (molar mass
204.2 g j mole).

This primary standard, KHP, is used to standardize the secondary standard, sodium hydroxide. The
standardized sodium hydroxide solution can then be used to determine the molar mass of an unknown acid
through titration.

In both steps a titration is performed in which a buret is used to dispense measured increments of the sodium
hydroxide solution into a second solution containing a known mass of KHP (NaOH standardization). For

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the second reaction, a mass of acid whose molecular weight is unknown is then titrated with the solution
of NaOH whose concentration was determined by the standardization with KHP. The stoichiometry of the
reaction depends on the number of ionizable hydrogens within the acid. KHP is a weak monoprotic acid
that will react with sodium hydroxide in a 1:1 mole ratio:

KHC 8 H A 4 + NaOH -> KNaC 8 H 4 A + H 2

The unknown acid may be monoprotic, diprotic, or triprotic dependent on the number of acidic hydrogens
present in the molecule. A monoprotic acid, HA, has one acidic hydrogen, a diprotic acid, H 2 A, two acidic
hydrogens, and a triprotic acid, H3A, three acidic hydrogens. The stoichiometries of the reactions are
shown below. You will be told whether your unknown acid is monoprotic, diprotic, or triprotic.

Monoprotic HA + NaOH -> NaA + H 2
Diprotic H 2 A + 2NaOH -^ Na 2 A + 2H 2
Triprotic H 3 A + ZNaOH -^ Na 3 A + 3H 2

The indicator phenolphthalein is used as a signal of the equivalence point. Phenolphthalein is a weak
organic acid that will change from colorless to pink near the equivalence point of the titrations. The actual
point at which the indicator changes color is the end point. The endpoint and the equivalence point are
not the same. The difference between the two is the titration error. Obviously, for a titration to be of
value, care must be taken to select an indicator for which the difference between the equivalence point and
the endpoint is small. With this particular titration, it is very small.

Pre-Lab Questions

1. Why is it necessary to rinse out the buret with the NaOH solution?

2. Calculate the approximate weight of KHP required so that about 40 mL of 0.1 M sodium hydroxide
is used in a titration. (MW of KHP = 204.23 g/mol)

3. Calculate the molarity of a KHP solution when 0.759 g of KHP is dissolved in 50.0 mL of water.

4. 0.521 g of KHP is dissolved in 40 mL of water, and titrated with a 0.102 M NaOH solution. Calculate
the number of mL of the NaOH solution added.

Purpose

The purpose of this experiment is to determine the concentration of a titrating solution, NaOH, using a
stable compound, KHP. Once the concentration of the NaOH solution is known, it can then be used to
determine the molecular weight of an acid whose formula is unknown.

Apparatus and Materials

• 50 mL buret

• buret stand

• buret clamp

• 125 mL Erlenmeyer flask

• phenolphthalein

• NaOH solution
. KHP, solid

• unknown acid, solid

• 10 mL graduate

• 400 mL beaker

• 100 mL graduate

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Safety Issues

NaOH is a caustic solution and will cause severe burns, especially to eye tissue. Wear goggles and aprons
at all times. The solid acids cause considerable irritation if exposed to skin or mucous membranes. Avoid
exposure.

Procedure for Part I

Part 1. Standardization of Sodium Hydroxide Solution

Obtain the primary acid standard from your instructor. Record the name of the acid, its molecular
formula, and number of acidic hydrogens per molecule. Prepare about 300 mL of approximately 0.1 M
sodium hydroxide by diluting 6 M NaOH with distilled water. (Calculate, ahead of time, how much water
and how much 6 M NaOH will be needed.)

WARNING: Concentrated sodium hydroxide is corrosive and causes severe burns. Handle with care.
Dilute and wash up spills with plenty of water. Wash affected skin with water until it no longer feels
slippery, but feels "squeaky" clean.

Store the solution in your plastic bottle, label it "0.1 M NaOH", and keep it tightly capped. You will
determine the exact molarity of this NaOH solution by standardization.

Calculate the mass of the primary acid standard that would react with about 20 mL of 0.1 M NaOH.
Weigh approximately this amount into a clean 125 - mL Erlenmeyer flask by taring the balance with the
flask on the pan, and then adding the acid to the flask. Record the mass of the primary acid standard
to the highest precision allowed by the electronic balance. Add 30 to 40 mL of your purified water to the
flask, and swirl to dissolve the primary acid standard. Add three or four drops of phenolphthalein indicator
solution to the flask, and swirl to mix well. Label this flask and keep it tightly capped until ready for use.

Rinse the inside of a CLEAN buret three times with small quantities of your 0.1 M sodium hydroxide
solution (called "rinsing in" with the solution to be used in the buret). Drain the rinses though the
stopcock and tip. Do not forget to rinse liquid through the tip, to replace water there. Fill the buret above
the 0.0 mL mark with 0.1 M NaOH, and then drain it until the meniscus is slightly below

NOTE: Do not waste the time it takes to set the starting level to exactly 0.0 mL. It is more efficient and
more accurate to set the level between 1 and 2 mL and read the starting level precisely.

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Tips on Technique

• To read the buret accurately, hold a white card with a black stripe behind the buret, with the black
stripe below the meniscus, and the meniscus itself in front of the white region above the black stripe
(see illustration). The meniscus will appear black against the white card. Keeping your eye level
with the meniscus, read the buret.

• Remember to estimate one more digit than those marked on the scale.

• Remember that the buret scale reads increasing volume downward, not upward.

Tips on Technique

• Record the starting level to 0.01 mL precision (as always, estimate one more digit than marks indicate) .

• Titrate the solution of primary acid standard with the 0.1 M NaOH until faint (see figure) phenolph-
thalein color appears and persists for 30 seconds. (Why might the color slowly disappear even after
all acid is titrated?) Record the final buret reading to 0.01 mL precision.

• Mix the solution in the titration flask thoroughly after each addition of titrant, to ensure complete
reaction before adding more.

• As you near the endpoint, wash the sides of the flask with distilled water to make sure that all
delivered titrant is in solution.

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When you see that you are within a drop or two of the endpoint, split drops to avoid overshooting
the endpoint.

Lab Procedure

Perform three titrations. For each, calculate the molarity of your NaOH solution. (From the mass of acid,
and its molecular weight, you can calculate the number of moles of acid, which is equal to the number of
moles of base you delivered. The molarity is found from the number of moles and the volume.)

When you have three values for the molarity of your NaOH solution, determine the average value.

Table 18.8: Data Table for Part I

Trial 1

Trial 2

Trial 3

Initial Reading, NaOH mL

buret

Final Reading, NaOH mL

buret

Volume of NaOH added mL

Grams of acid standard g

Moles of acid standard mol

Molarity of NaOH M

mL
mL

g
mol

mL
mL

g
mol

Average molarity of NaOH = M

Procedure for Part II

Part 2. Finding the Molar Mass of an Unknown Acid

• Obtain a sample of a solid unknown from your instructor. Record its ID code in your report.

• Also, record the approximate amount of unknown to use in each titration, and the number of acid
hydrogens per molecule. Your instructor will provide this information.

• Weigh the suggested amount into a clean 125 - mL erlenmeyer flask, by taring the balance with the
flask on the pan, and then adding the acid to the flask. Record the mass of the sample to the precision
allowed by the balance. Add 30 to 40 mL of distilled water and swirl to dissolve your sample. Add
three to four drops of phenolphthalein indicator solution and swirl to mix well.

• Titrate your sample with your standardized NaOH solution until faint phenophthalein color persists
for 30 seconds. *If your titration requires 10 to 25 mL of NaOH solution, carry out a second titration
with an unknown sample of about the same mass. Otherwise, adjust the sample mass to bring the
expected end-point volume to between 10 and 25 mL and do two more titrations.

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• For each titration, compute the molar mass of the unknown acid, to the precision allowed by your
data. Do three titrations and report the average molar mass for the solid acid.

Data for Part II

ID code of acid

Number of Acidic Hydrogens in Acid
Approximate mass of acid to be used

Table 18.9: Data Table for Part II

Trial 1

Trial 2

Trial 3

Mass of unknown acid g
sample

Volume of NaOH used mL

Mols of NaOH used moL

Moles of acid present 1 mol

Molar mass of acid 2 g/mol

Molarity of NaOH M

mL
moL
mol
g/mol

1 Moles of acid present

2 Molar mass of acid
Average Molar Mass

moles NaOH

number of H + ions per acid molecule

2 Molar mass of acid = g ' a " li acu

moles acid

_g/mol

18.4 Buffers

Lesson Objectives

• Define a buffer and give various examples of buffers.

• Explain the effect of a strong acid on the pH of a weak acid/conjugate base buffer.

• Explain the effect of a strong base on the pH of a weak base/conjugate acid buffer.

Introduction

There are many situations in which it is desirable to keep the pH of a solution close to a particular value
even though quantities of acids and/or bases are added to the solution. Many organic and biochemical
reactions require acids or bases in the reaction but if the pH goes too high or too low, the products will be
destroyed. For these reactions, it is necessary to keep the pH within a very small range even while acids
or bases are added to the reaction. Chemists use mixtures called buffers to keep the reaction solutions
within the necessary pH range. Buffers are mixtures of chemicals that cause a solution to resist changes
in pH.

Buffers are very important to many biological reactions. Human blood is a substance whose function is
very dependent on the function of buffers. Human blood must maintain an almost constant pH between 7.3
and 7.5 . If a person's blood pH goes 0.2 outside the acceptable range, the person will become unconscious
and the blood pH goes 0.4 outside the range, the person will die. The pH of human blood can change
depending on foods we eat and the rate at which we inhale and exhale C02- Fortunately, the human blood

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stream has buffers which are able to resist pH changes due to food intake and breathing rates.

Buffers

A buffer is a solution that maintains the pH level when small amounts of acid or base are added to the
system. Buffer solutions contain either a weak acid and the conjugate base of the weak acid or a weak
base and the conjugate acid of the weak base. A common procedure for producing a buffer in the lab is
to make a solution of a weak acid and a salt of that weak acid. For example, you could make a buffer by
making a solution containing acetic acid and sodium acetate.

The common buffer mentioned above would contain acetic acid, CH3COOH, and the acetate ion, CHj,COO~ .
This buffer can keep a solution in the range of a pH of 3.7 - 5.8 even though small amounts of acids or
bases are added to the solution. Another example is hydrogen phosphate ion, HPO^~, and the phosphate
ion, PO^~ which will buffer a solution in the pH range of 11.3 - 13.3.

How is it possible that a solution will not change its pH when an acid or base is added? Let's examine the
acetic acid/acetate ion buffer. The ionization equation for acetic acid is shown below.

HC2H S 2 ( cl q) £+ H + C 2 H 3 2

It is obvious that if acid {H + ions) is added to this solution, the equilibrium will shift toward the reactants
to use up some of the added hydrogen ions. Equilibrium will be re-established in the solution with different
concentrations of the three species in the reaction. It is also obvious that if a base is added to this solution,
the base will remove hydrogen ions and the equilibrium will shift to the right to partially counteract the
stress. Again, equilibrium will be re-established with new concentrations.

What is not obvious about this buffer solution is that acetic acid is a weak acid and therefore, most of
the acid dissolved in the solution remain as acid molecules and do not dissociate. Therefore, there will a
large quantity of undissociated HC 2 H^0 2 . The fact that we also dissolved sodium acetate in this solution
provides a large quantity of acetate ions, CiH^O^ , in the solution. The existence of large quantities of both
undissociated acid molecules and acetate ions in the solution is what allows the buffer to consume quite a
large amount of added acid or base without the pH changing significantly.

Examine what happens to 1.00 liter of pure water to which 0.100 mole of gaseous HCl is added.

The original concentration of hydrogen ion in the pure water is 1.00 x 10~ 7 M and therefore, the pH is 7.

After the 0.100 mole of HCl is added, the concentration of the hydrogen ion will be 0.100 M (plus the
original 1.00 x 10~ 7 M which can be neglected as not significant). This new concentration of hydrogen ion
will produce a pH = 1. So, the addition of the 0.100 mole of gaseous HCl caused the pH of the pure water
to change from 7 to 1.

Let's now see what happens if this same amount of gaseous HCl is added to an acetic acid-acetate ion
buffer. We'll the same amount of original solution, 1.00 liter, and let's say we made this solution to contain
0.50 M acetic acid and 0.50 M acetate ion (0.50 M sodium acetate which totally dissociated). The acetic
acid dissociation equation will reach equilibrium and its K a will be 1.8 x 10 -5 .

[HC 2 H 3 2 ]

We know both the acetic acid and the acetate ion concentrations will be 0.50 M, so we can plug these
values into the expression and solve for [H + ].

= (1.8X10-» 3 2] = (1.8 X 10^)(0.50) =
1 ' [C2H3O5] (0.50)

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Then we can insert the hydrogen ion concentration into the pH formula and determine the original pH of
the buffer solution.

pH = -log(1.8 x 1(T 5 ) = 4.74

Next, we will add the same 0.100 mole of gaseous HCl to this buffer solution and calculate the pH of the
solution after the acid has been added and equilibrium has been re-established.

When we add 0.100 mole HCl gas to this solution, the added hydrogen ion will combine with acetate ion
to produce more undissociated acid. A small amount of the added hydrogen ions may remain as ions but
the amount is, once again, beyond the significant figures of the problem and can be neglected. The new
[HC 2 H 3 2 ] will equal 0.60 M (the original 0.50 M plus the added 0.10 M) and the new [C 2 H 3 2 ~] will
equal 0.40M (the original 0.50 M minus the 0.10 M that reacted with the added hydrogen ions). We can
now plug these values into the K a expression, calculate the new [H + ], and find the new pH.

[H+][C 2 H 3 2 ]

K a = -r-^ f- = 1.8 x 10~ 5

[HC 2 H 3 2 ]

r .. (1.8xl0- 5 )[//C 2 // 3 O 2 ] (1.8xl0- 5 )(0.60)

\H + ] = - = J = - £ - = 2.7 x 10~ 5 M

L J [C 2 H 3 0~] (0.40)

pH = -log (2.7 x 10~ 5 ) = 4.57

The same quantity of HCl gas that changed the pH of pure water from 7 to 1 has changed the pH of this
buffer from 4.74 to 4.57. . . only a change of 0.17. . . that is the function of a buffer. Buffers resist change
to pH. We could this same calculation but add a base instead of an acid and show that the pH increases
by this same slight amount. Maybe that would be good practice for you.

Sample question: Which of the following combinations would you expect possible to make into buffer
solutions:

(a) HCl0 4 /ClO A -

(b) CH 3 NH 2 /CH 3 NH 3 +
Solution:

(a) HCIO4/CIO4 : HCIO4 is a strong acid and buffers are made from weak acids and their conjugate bases
or weak bases and their conjugate acids. Therefore this cannot be made into a buffer solution.

(b) CH 3 NH 2 /CH 3 NH 3 + : CH 3 NH 2 is a weak base and CH 3 NH 3 + is the conjugate acid of this base. Therefore
this can be made into a buffer solution.

The Buffer in Blood

The primary buffer found in your bloodstream is carbonic acid, H 2 C0 3 . The carbonic acid is present due
to carbon dioxide from your respiratory system dissolving in water.

C0 2{g) + H 2 {L) i? H 2 C0 3{aq) Equation 1

The amount of carbonic acid in your bloodstream is affected by the rate of your respiration. If you breathe
rapidly, you reduce the amount of C0 2 in your bloodstream and the equilibrium shown in Equation 1 shifts
toward the reactants thus lowering the amount of H 2 C0 3 . If you breathe slowly, the amount of C0 2 in
your bloodstream increases and the equilibrium in Equation 1 shifts toward the products increasing the
amount of H 2 C0 3 in your system.

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Once the H2CO3 is produced by dissolving carbon dioxide, the carbonic acid dissociates in your blood as
shown in Equation 2.

H2C0 3{aq) <=> H + + HC0 3 Equation 2

This is the buffering reaction in your blood, composed of the weak acid, H2CO3, and its conjugate base,
HC0 3 .

Any changes in blood pH that could be caused by food intake, will be buffered by this equilibrium system.
If acid is added to your blood, the equilibrium in Equation 2 will shift toward the reactants using up the
hydrogen ions and if base were added to your blood thus reacting with hydrogen ions, the equilibrium will
shift toward the products generating more hydrogen ions. This buffer in your blood is very efficient at
keeping your blood pH in the necessary range.

A problem exists for a few people because their respiratory rate is highly affected by their emotional
state. Some people, when they get nervous, begin breathing very fast or very slow. Breathing too fast is
called hyperventilating and breathing too slow is called hypoventilating. Your respiratory rate is normally
controlled by the amount of carbon dioxide in the blood. Your body receives instructions to breathe faster
or slower to adjust the amount of carbon dioxide in your blood in order to properly regulate the buffer
system. When people people breathe too fast or too slow because of other reasons, the CO2 content of the
blood becomes incorrect and the pH of the blood rises or lowers outside the acceptable range of 7.3 - 7.5
. When this happens, the person passes out or becomes unconscious. People who hyperventilate when
excited or nervous are sometimes advised to carry a lunch sack or something similar to breathe into when
they are feeling light-headed. Breathing into a sack returns air with the same concentration of carbon
dioxide that was exhaled . . . this keeps the amount of carbon dioxide in the blood up and avoids passing
out.

Lesson Summary

• A buffer is a solution of a weak acid and its conjugate base or a weak base and its conjugate acid
that resists changes in pH when an acid or base is added to it.

• Adding a strong acid to a weak acid/conjugate base buffer only decreases the pH by a small amount.

• Adding a strong base to a weak base/conjugate acid buffer only increases the pH by a small amount.

Review Questions

1. Define a buffer solution.

2. What are two different types of buffer solutions?

3. One of the following statements of buffers is incorrect. Which one?

(a) A buffer may be prepared from a weak acid and its conjugate base salt.

(b) A buffer may be prepared from a weak base and its conjugate acid salt.

(d) A buffer is a solution that can resist changes in pH when a small amount of acid or base is
added to it.

4. Which pair of aqueous 1.0 mol/L solutions could be chosen to prepare a buffer?

5. (a) i and iii only

(b) i and iii only

(d) None of these solutions is a buffer.

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i. NH^HSO^aq) and H 2 SO^ aq)
h. HN0 2(aq) and NaN0 2{aq)
hi. NH^Cl[a q ) and NH^ aq)

6. Which of the following would form a buffer solution if combined in appropriate amounts?

(a) HCl and NaCl

(b) HCN and NaCAT

(d) HN0 3 and AWV0 3

7. A buffer is made up of a weak acid and a conjugate base. A small amount of acid is added to the
buffer. What happens to the resulting solution?

(a) The acid dissociation constant goes up.

(b) The concentration of the weak acid in the buffer goes down.

(d) The pH remains almost the same.

8. Almonds from the wild have a very bitter taste because of hydrogen cyanide (and therefore are very
dangerous to eat!!!). Interestingly if we think about HCN in a buffer situation, HCN and NaCN can
be considered to act as a buffer solution. Sulfurous acid is used quite frequently as a cleansing agent.
If we take the sodium salt, Na 2 SOs, of sulfurous acid, H 2 SOs we do not make a buffer solution. Why
is this so? Why would one make a buffer solution and not the other?

Further Reading / Supplemental Links

• http://en.wikipedia.org

Vocabulary

buffer A buffer is a solution of a weak acid and its conjugate base or a weak base and its conjugate acid
that resists changes in pH when an acid or base is added to it.

Image Sources

(2
(3
(4
(5
(6
(7
(8
(9

Hydrangeas. Public Domain.
Therese Forsythe. . CC-BY-SA.

Theresa Knott. Typical Titration Setup.. GNU Free Documentation License.
Datamax. An electronic pH meter. Public Domain.
Andel Fruh. Structure of Phenolphthalein.. Public Domain.

Structure of Methyl Orange (Yellow Form).. Public Domain.
Therese Forsythe. Titration Curve for a Strong Acid vs. Weak Base.. CC-BY-SA.
Therese Forsythe. Titration Curve for a Strong Acid vs. Strong Base.. CC-BY-SA.
Richard Parsons. The color of Universal Indicator at various pH's.. CC-BY-SA.

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Chapter 19

Radioactivity and the Nucleus

19.1 Discovery of Radioactivity

Lesson Objectives

The student will:

• describe the roles played by Henri Becquerel and Marie Curie played in the discovery of radioactivity.

• list the most common emissions from naturally radioactive nuclei.

Introduction

No one could have known in the 1800s that the discovery of the fascinating science and art form of
photography would eventually lead to the splitting of the atom. The basis of photography is the fact
that visible light causes certain chemical reactions. If the chemicals are spread thinly on a surface but
protected from light by a covering, no reaction occurs. When the covering is removed, however, light
acting on the chemicals causes them to darken. With millions of cameras in use today we do not think
of it as a strange phenomenon, but at the time of its discovery photography was a strange and wonderful
thing. Even stranger was the discovery by Roentgen, that radiation other than visible light could expose
photographic film. He found that film wrapped in dark paper would react when x-rays went through the
paper and struck the film.

Becquerel and Radioactivity

When Becquerel heard about Roentgen's discovery, he wondered if his fluorescent minerals would give
the same x-rays. Becquerel placed some of his rock crystals on top of a well-covered photographic plate
and sat them in the sunlight. The sunlight made the crystals glow with a bright fluorescent light, but
when Becquerel developed the film he was very disappointed. He found that only one of his minerals, a
uranium salt, had fogged the photographic plate. He decided to try again, and this time, to leave them
out in the sun for a longer period of time. Fortunately, the weather didn't cooperate and Becquerel had to
leave the crystals and film stored in a drawer for several cloudy days. Before continuing his experiments,
Becquerel decided to check one of the photographic plates to make sure the chemicals were still good. To
his amazement, he found that the plate had been exposed in spots where it had been near the uranium
containing rocks and some of these rocks had not been exposed to sunlight at all.

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Bequerel's radioactive rock sitting on an envelope containing photographic film.

Rock resting on covered film.

Shadow on developed film.

In later experiments, Becquerel confirmed that the radiation from the uranium had no connection with
light or fluorescence, but the amount of radiation was directly proportional to the concentration of uranium
in the rock. Becqueral had discovered radioactivity.

The Curies and Radium

One of Becquerel's assistants, a young Polish scientist named Maria Sklowdowska (to become Marie Curie
after she married Pierre Curie), became interested in the phenomenon of radioactivity. With her husband,
she decided to find out if chemicals other than uranium were radioactive. The Austrian government was
happy to send the Curies a ton of pitchblende from the mining region of Joachimstahl because it was waste
material that had to be disposed of anyway. The Curies wanted the pitchblend because it was the residue of
uranium mining. From the ton of pitchblend, the Curies separated 0.10 g of a previously unknown element,
radium, in the form of the compound, radium chloride. This radium was many times more radioactive
than uranium.

By 1902, the world was aware of a new phenomenon called radioactivity and of new elements which
exhibited natural radioactivity. For this work, Becquerel and the Curies shared the 1903 Nobel Prize in
Physics. For her subsequent work in radioactivity, Marie Curie won the 1911 Nobel Prize in Chemistry.
She was the first female Nobel laureate and the only person ever to receive two Nobel Prizes in two different
scientific categories.

Marie Sklodowska before she moved to Paris. (Source: http : //chemistry . about . com/od/historyof chemistr
ig/Pictures-of-Famous-Chemists/Marie. --bs.htm. Public Domain)

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Further experiments provided information about the characteristics of the penetrating emissions from
radioactive substances. It was soon discovered that there were three common types of radioactive emissions.
Some of the radiation could pass easily through aluminum foil while some of the radiation was stopped by
the foil. Some of the radiation could even pass through foil up to a centimeter thick. The three basic types
of radiation were named alpha, beta, and gamma radiation. The actual composition of the three types of
radiation was still not known.

Eventually, scientists were able to demonstrate experimentally that the alpha particle, a, was a helium
nucleus (a particle containing two protons and two neutrons), a beta particle, /?, was a high speed electron,
and gamma rays, y, were a very high energy form of light (even higher energy than x-rays).

Lesson Summary

• Henri Becquerral, Marie Curie, and Pierre Curie shared the discovery of radioactivity.

• The most common emissions of radioactive elements were called alpha (a), beta (/?), and gamma (y)
radiation.

Vocabulary

alpha particle An alpha particle is a helium-4 nucleus.

beta particle A beta particle is a high speed electron, specifically an electron of nuclear origin.

gamma ray Gamma radiation is the highest energy on the spectrum of electromagnetic radiation.

Marie Curie Marie Curie was a physicist and chemist of Polish upbringing, and subsequently, French
citizenship; a pioneer in the field of radioactivity and the only person to ever win two Nobel prizes
in science.

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Further Reading / Supplementary Links

http : //www. chem . duke . edu/~ jds/cruise_chem/nuclear/discovery . html

Review Questions

1. Put the letter of the matching phrase on the line preceding the number. 1. alpha particle - a.

high energy electromagnetic radiation

2. beta particle - b. a high speed electron

3. gamma ray - c. a helium nucleus

19.2 Nuclear Notation

Lesson Objectives

The student will:

• state the information contained in the atomic number of a nucleus.

• state the information contained in the mass number of a nucleus.

• subtract the atomic number from the mass number to determine the number of neutrons in a nucleus.

• read and write complete nuclear symbols (know the structure of the symbols and understand the
information contained in them).

Introduction

Just as chemical formulas use symbols and chemical equations use symbols, nuclei are represented by
symbols. The complete nuclear symbol contains the symbol for the element and numbers that relate to
the number of protons and neutrons in that particular nucleus.

Atomic and Mass Numbers

The identity of an atom is determined by the number of protons in the nucleus of the atom. The number
of protons in the nucleus of the atom is also known as the ATOMIC NUMBER. The atomic number is
the smaller number appearing on the periodic table for each atom. The atomic number for hydrogen is 1.
This means that if a nucleus has 1 proton in it, it is a hydrogen nucleus no matter how many neutrons it
has. The mass number for a nucleus is the total number of protons and neutrons (nucleons) in the nucleus
of an atom. Both the mass number and the atomic number for nuclei are always whole numbers because
there are no fractions of nucleons. To find the number of neutrons in the nucleus, you would subtract the
atomic number from the mass number. For example, if the atomic number for a nucleus was 8 and the
mass number was 18, the nucleus would contain 8 protons (equal to the atomic number) and 10 neutrons
(18 nucleons - 8 protons = 10 neutrons).

The Complete Nuclear Symbol

To write a complete nuclear symbol, the mass number is placed at the upper left (superscript) of the
chemical symbol and the atomic number is placed at the lower left (subscript) of the symbol. The complete
nuclear symbol for helium-4 is drawn below.

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The complete nuclear symbol for helium-4. (Source: Richard Parsons. CC-BY-SA)

mass number —

_ _ _ chemical symbol

atomic number

The following nuclear symbols are for a nickel nucleus with 31 neutrons and a uranium nucleus with 146
neutrons.

59 l\Ti 238 j j

28 i>x 92 *-*

In the nickel nucleus represented above, the atomic number 28 indicates the nucleus contains 28 protons,
and therefore, it must contain 31 neutrons in order to have a mass number of 59. The uranium nucleus
has 92 protons as do all uranium nuclei and this particular uranium nucleus has 146 neutrons. Another
way of representing these nuclei would be Ni - 59 and U - 238.

Dalton's original atomic theory stated that all atoms of an element were identical in every way, but it was
later discovered by Thomson that atoms of an element were identical in every way except they could have
different mass numbers. This mass difference results from nuclei of the same element having a different
number of neutrons. In order to be the same element, nuclei must have the same number of protons, but
they may have a different number of neutrons. Atoms with the same atomic number but a different mass
number are called isotopes.

1
1

H *h :h

hydrogen- 1 hydrogen-2 hydrogen-3

(protium) (deuterium) (tritium)

Hydrogen, for example, has three isotopes, shown in the figure above. All three of hydrogen's isotopes
must have one proton (to be hydrogen), but they have zero, one, or two neutrons in the nucleus.

Originally, the names protium, deuterium, and tritium were suggested for the three isotopes of hydrogen,
but about a year after the names were suggested it was felt that a new name indicated a new element,
which these were not. Furthermore, naming compounds using these names would cause great difficulty.
Nuclear scientists today use the names hydrogen-1, hydrogen-2, and hydrogen-3, but occasionally you
will see or hear the other names.

Lesson Summary

• The complete nuclear symbol has the atomic number (number of protons) of the nucleus as a subscript
at the lower left of the chemical symbol and the mass number (numberofprotons + neutrons) as a
superscript at the upper left of the chemical symbol.

Vocabulary

atomic number The atomic number indicates the number of protons in the nucleus.
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mass number The mass number indicates the number of protons plus the number of neutrons in the
nucleus.

electron An electron is a fundamental sub-atomic particle that carries a negative charge.

neutron A neutron is a sub-atomic particle with no electric charge and a mass slightly larger than a
proton.

proton A proton is a fundamental sub-atomic particle with a net positive charge.

nucleus The nucleus of an atom if the very dense region, consisting of nucleons (proton and neutrons)
at the center of an atom.

nuclei Nuclei is the plural of nucleus.

nucleon A nucleon is a constituent part (proton or neutron) of an atomic nucleus.

nuclide A type of nucleus specified by its atomic number and mass number.

Review Questions

1. Write the complete nuclear symbol for a nucleus of chlorine that contains 17 protons and 20 neutrons.

2. Write the complete nuclear symbol for a nucleus of oxygen that contains 8 protons and 10 neutrons.

3. If a nucleus of uranium has a mass number of 238, how many neutrons does it contain?

4. In the nuclear symbol for a beta particle, what is the atomic number?

5. Is it possible for isotopes to be atoms of different elements? Explain why or why not.

6. How many neutrons are present in a nucleus whose atomic number is one and whose mass number is
one?

7. Name the element of an isotope whose mass number is 206 and whose atomic number is 82.

8. How many protons and how many neutrons are present in a nucleus of lithium- 7?

9. What is the physical difference between at/- 235 atom and at/- 238 atom?
10. What is the difference in the chemistry of a U — 235 atom and at/- 238 atom?

19.3 Nuclear Force

Lesson Objectives

The student will:

compare the energy released per gram of matter in nuclear reactions to that in chemical reactions,
express the equation for calculating the change in mass during nuclear reactions that is converted
into energy,
express the relationship between nuclear stability and the nuclei's binding energy per nucleon ratio.

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Introduction

There are only four forces in nature that produce all interactions between objects. Gravity affects all
particles and only attracts, never repels. It is the weakest of the four forces but acts over great distances.
The electromagnetic force acts between electric charges and magnetic fields and causes all physical and
chemical processes to occur. It can also act at large distances. The weak nuclear force is limited to the
atomic nucleus and causes unstable particles and nuclei to decay. The strong nuclear force is also limited
to the nucleus and binds quarks into nucleons and nucleons into nuclei.

Nuclear Force Overcomes Proton Repulsion

A nucleus (with one exception, hydrogen-1) consists of some number of protons and neutrons pulled
together in an extremely tiny volume. Since protons are positively charged and like charges repel, it is
clear that protons cannot remain together in the nucleus unless there is a powerful force holding them there.
The force which holds the nucleus together is generated by nuclear binding energy. We are concerned not
with just the total amount of binding energy a nucleus possesses, but also with how many nucleons (protons
and neutrons) are present for the binding energy to hold together. It is more illuminating for us to consider
the amount of binding energy per nucleon that a nucleus contains. A nucleus with a large amount of
binding energy per nucleon will be held together tightly and is referred to as stable. When there is too
little binding energy per nucleon, the nucleus will be less stable and may disintegrate (come apart). When
nuclei come apart, they come apart violently accompanied by a tremendous release of energy in the form
of heat, light, and radiation.

Mass Defect Becomes Binding Energy

We know that a proton has a mass of 1.00728 Daltons (also known as atomic mass units) and a neutron
has a mass of 1.00867 Daltons. A helium-4 nucleus consists of two protons and two neutrons. The mass
of these separate protons and neutrons would be:

2 x 1.00728 + 2 x 1.00867 = 4.03190 Daltons.

The actual mass of a helium-4 nucleus is known to be 4.00150 Daltons. It would appear that some mass
has been lost when the particles formed a nucleus. This difference between the sum of the masses of the
individual nucleons and the mass of the corresponding nucleus always occurs and has been called mass
defect. The mass defect in this case is 0.03040 Daltons. This mass, of course, is not lost but is converted
into energy - binding energy. Albert Einstein first theorized that mass and energy could be converted into
one another and he produced the equation with which to calculate the conversion; E = mc 2 , where E is
energy in Joules, m is the mass in kilograms, and c is the speed of light, 3 x 10 8 meters/second. If we think
about the conservation of mass and energy, we can account for everything. Nothing is lost and nothing
appears from nowhere. The conversion of a very small amount of mass into energy produces an immense
amount of energy, as you might guess from the size of the number you get when you square the speed of
light. If 1.00 gram of mass were converted to energy in this manner, it would produce 9.0 x 10 13 Joules.
This amount of energy would raise the temperature of 300 million liters of water from room temperature
to boiling.

More Binding Energy per Nucleon Produces Stability

It is conventional to plot the binding energy per nucleon (total binding energy divided by the number of
nucleons in the nucleus) versus the atomic mass of the nucleus. Such a graph is shown in Figure 19.1. The

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position of greatest binding energy per nucleon is held by iron-56. Nuclei both larger and smaller than
iron-56 have less binding energy per nucleon and are therefore, less stable. In the graph below, binding
energy is measured in Mev (million electron volts). One million electron volts is equal to 1.6 x 10 -13 Joules.
(This graph has been smoothed. The actual graph line zig zags up and down a little.)

c ,

^Fe

o '

23S,,

~— — — —^ iL

f—C

NHe

IXI

1 H

Mass Number

Figure 19.1: The graph of binding energy per nucleon for atoms between atomic number 1 and 92.

Lesson Summary

• The proton-proton repulsion in a nucleus is overcome by binding energy to hold the nucleus together.

• The sum of the masses of the individual components of a nucleus is greater than the mass of the
nucleus and the "lost" mass is called mass defect.

• Much of the mass defect is converted into binding energy according the Einstein equation, E

• The stability of a nucleus is determined by the amount of binding energy present for each nucleon

• Nuclei with lower binding energy per nucleon may disintegrate.

• The nucleus with the greatest binding energy per nucleon is iron-56.

mc .

Vocabulary

binding energy Binding energy is the amount of energy that holds a nucleus together, and therefore,
also the amount of energy required to decompose a nucleus into its component nucleons.

mass defect Mass defect is the difference between the sum of the masses of the nuclear components and
the mass of the corresponding nucleus. Much of this lost mass is converted into binding energy.

nucleon Nucleon is a collective name for neutrons and protons.

Further Reading / Supplementary Links

http://www.wisegeek.com/what-is-the-strong-nuclear-force.htm

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Review Questions

1. Iron-56 is a very stable nucleus while cobalt-60 is an unstable nucleus. Which nucleus would you
expect to have more binding energy per nucleon?

2. Calculate the mass defect and binding energy for a mole of carbon- 14 given the data below.

The molar mass of carbon- 14 is 14.003241 g/mol.

The molar mass of a proton is 1.007825 g/mol.

The molar mass of a neutron is 1.008665 g/mol.

Mass in kilograms is converted into energy in Joules by multiplying the mass times the speed of light
squared, E = mc 2 .

The speed of light is 3.00 x 10 8 m/s.

kg • m 2 /s 2 = Joules

19.4 Nuclear Disintegration

Lesson Objectives

The student will:

• list some naturally occurring isotopes of elements that are radioactive.

• describe the three most common emissions during natural nuclear decay.

• express the changes in the atomic number and mass number of a radioactive nuclei when an alpha
particle is emitted.

• express the changes in the atomic number and mass number of a radioactive nuclei when a beta
particle is emitted.

• express the changes in the atomic number and mass number of a radioactive nuclei when a gamma
ray is emitted.

• express that protons and neutrons are not indivisible and are composed of particles called quarks.

• express the number of quarks that make up a proton or neutron.

Introduction

Under certain conditions, less stable nuclei alter their structure to become more stable nuclei. Those
unstable nuclei that are larger nuclei than iron-56 spontaneously disintegrate by ejecting particles. This
process of decomposing to form a different nucleus is called radioactive decay. Many nuclei are radioactive;
that is, they decompose by emitting particles and in doing so, become a different nucleus. All nuclei with
84 or more protons are radioactive and many elements with less than 84 protons have both stable and
unstable isotopes.

Types of Radioactive Decay

In natural radioactive decay, three common emissions occur. When these emissions were originally ob-
served, scientists were unable to identify them as some already known particle and so named them alpha
particles (a), beta particles (/?), and gamma rays (y). At some later time, alpha particles were
identified as helium-4 nuclei, beta particles were identified as electrons, and gamma rays as a form of

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electromagnetic radiation like x-rays except much higher in energy and even more dangerous to living
systems.

Alpha Decay

The nuclear disintegration process that emits alpha particles is called alpha decay An example of a nucleus
that undergoes alpha decay is uranium-238. The alpha decay of U - 238 is

In nuclear equations, it is required that the sum of the atomic numbers on the reactant side equal the sum
of the atomic numbers on the product side and the same is true for the mass numbers on the two sides.
In this equation,

atomic number : 92 = 2 + 90

mass number: 238 = 4 + 234.

Therefore, the equation is balanced.

Another alpha particle producer is thorium-230.

230 ,

4„

226„

90 Th

2 He +

88 Ra

Confirm that this equation is correctly balanced.

Beta Decay

Another common decay process is beta particle production, or beta decay. It may occur to you that we
have a logically difficult situation here. Nuclei do not contain electrons and yet during beta decay, an
electron is emitted from a nucleus. At the same time that the electron is being ejected from the nucleus,
a neutron is becoming a proton. It is tempting to picture this as a neutron breaking into two pieces with
the pieces being a proton and an electron. That would be convenient for simplicity, but unfortunately that
is not what happens; more about this at the end of this section.

In order to insert an electron into a nuclear equation and have the numbers add up properly, an atomic
number and a mass number had to be assigned to an electron. The mass number assigned to an electron is
zero (0) which is reasonable since the mass number is the number of protons plus neutrons and an electron
contains no protons and no neutrons. The atomic number assigned to an electron is negative one (-1)
because that allows a nuclear equation containing an electron to balance atomic numbers. Therefore, the
nuclear symbol representing an electron (beta particle) is

°«

_i e or ,\P

Thorium-234 is a nucleus that undergoes beta decay. Here is the nuclear equation for this beta decay.

234 ml 234„

90 Th -* -I e + 91 Pa

Note that both the mass numbers and the atomic numbers add up properly:

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atomic number: 90 = -1 + 91

mass number: 234 = + 234

The mass numbers of the original nucleus and the new nucleus are the same because a neutron has been
lost, but a proton has been gained and so the sum of protons plus neutrons remains the same. The
atomic number in the process has been increased by one since the new nucleus has one more proton than
the original nucleus. In this beta decay, a thorium-234 nucleus has become a protactinium-234 nucleus.
Protactinium-234 is also a beta emitter and produces uranium-234.

234„

234 XT

^ Pa

— >

„e

™ u

-1

Once again, the atomic number increases by one and the mass number remains the same; confirm that the
equation is correctly balanced.

Protons and Neutrons are Made Up of Quarks

Protons and neutrons are not fundamental particles as electrons are. Protons and neutrons are composed
of more fundamental particles called quarks. Quarks are fundamental particles not similar to electrons,
but like electrons in that they have no smaller pieces. (If you think "quarks" is a somewhat fanciful name,
the world of particle physics holds a lot of surprises for you.) Quarks come in six different types (particle
physicists call them flavors). The six "flavors" of quarks are named up, down, top, bottom, strange, and
charmed. Protons and neutrons are composed of a combination of up and down quarks. The up quark
carries a charge of +2/3 and the down quark carries a charge of -1/3. A proton is composed of two up
quarks and one down quark. Thus the charge on a proton is +2/3 + 2/3 - 1/3 = 1. A neutron is composed
of one up quark and two down quarks, hence its charge is +2/3 - 1/3 - 1/3 = 0. With this information,
you can see that a neutron is NOT composed of a proton and an electron. The beta particle produced
during beta decay is created in the process of a neutron decaying to a proton. We will view the process in
terms of the net effect which is changing a neutron into a proton and emitting an electron.

Gamma Radiation

Frequently, gamma ray production accompanies nuclear reactions of all types. In the alpha decay of £7-238,
two gamma rays of different energies are emitted in addition to the alpha particle.

238 TT 4 TT 234^, n

92 U ^2 He+ 90 Th+ V

Virtually all of the nuclear reactions in this chapter also emit gamma rays, but for simplicity the gamma
rays are generally not shown. Nuclear reactions produce a great deal more energy than chemical reactions.
Chemical reactions release the difference between the chemical bond energy of the reactants and products,
and the energies released have an order of magnitude of 1 x 10 3 kJ/mol. Nuclear reactions release some of
the binding energy and may convert tiny amounts of matter into energy. The energy released in a nuclear
reaction has an order of magnitude of 1 x 10 8 kJ/mol.

Decay Series

The decay of a radioactive nucleus is a move toward becoming stable. Often, a radioactive nucleus cannot
reach a stable state through a single decay. In such cases, a series of decays will occur until a stable nucleus

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is formed. The decay of U — 238 is an example of this (Table 19.1). The U - 238 decay series starts with

U — 238 and goes through fourteen separate decays to finally reach a stable nucleus, Pb — 206. There are

similar decay series for U — 235 and Th — 232. The U — 235 series ends with Pb — 207 and the Th — 232 series
ends with Pb - 208.

Table 19.1:

Nucleus Type of Decay Product

U - 238 a Th- 234

Th - 234 B Pa- 234

Pa - 234 B U - 234

U - 234 a Th- 230

Th - 230 a Ra - 226

Ra - 226 a Rn - 222

Rn - 222 a Po - 218

Po- 218 a Pb- 214

P»-214 /? Bi -214

5/ -214 /? Po-214

Po-214 a P6-210

P6-210 /? Bi -210

5/ -210 /? Po-210

Po - 210 a- Pb - 206

Several of the radioactive nuclei that are found in nature are present there because they are produced in
one of the radioactive decay series. That is to say, there may have been radon on the earth at the time of
its formation, but that original radon would have all decayed by this time. The radon that is present now
is present because it was formed in a decay series.

Lesson Summary

• A nuclear reaction is one that changes the structure of the nucleus of an atom.

• The atomic numbers and mass numbers in a nuclear equation must be balanced.

• Protons and neutrons are made up of quarks.

• The two most common modes of natural radioactivity are alpha decay and beta decay.

• Most nuclear reactions emit energy in the form of gamma rays.

Vocabulary

alpha decay Alpha decay is a common mode of radioactive decay in which a nucleus emits an alpha
particle (a helium-4 nucleus).

beta decay Beta decay is a common mode of radioactive decay in which a nucleus emits beta particles.
The daughter nucleus will have a higher atomic number than the original nucleus.

quark Quarks are physical particles that form one of the two basic constituents of matter. Various species
of quarks combine in specific ways to form protons and neutrons, in each case taking exactly three
quarks to make the composite particle.

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Further Reading / Supplementary Links

Video showing nuclear reactions.

http : //www. youtube . com/watch?v=_f 4kYYAQC3c

Short animation showing the bending of a, ft, and y radiation under the influence of electrically charged
plates.

http : //www.mhhe . com/physsci/chemistry/chemistryessential/f lash/radioa7 . swf

Review Questions

1. Write the nuclear equation for the alpha decay of radon- 198.

2. Write the nuclear equation for the beta decay of uranium-237.

3. There are six known quarks. The experimenter who discovers particles in nuclear physics gets the
right to name the new particle. This has resulted in some very fanciful names for quarks. The six
quarks are named, up quarks, down quarks, charmed quarks, strange quarks, top quarks, and
bottom quarks. (The top and bottom quarks were originally named truth and beauty quarks, but
the names were changed for some reason.) Protons and neutrons are each made of only up and
down quarks and they are made of three quarks each. The up quark carries a charge of +o and the
down quark carries a charge of — g. Determine by the final charge on the proton and neutron, what
combination of three up and down quarks are required to make a proton and what combination will
make a neutron?

19.5 Nuclear Equations

Lesson Objectives

The student will:

• define and give examples of fission and fusion.

• classify nuclear reactions as fission or fusion.

• correctly fill in the missing particle in a nuclear equation with one species missing

• write balanced equations for nuclear transmutations.

Introduction

Atomic nuclei with an inadequate amount of binding energy per particle are unstable and occasionally
disintegrate in an organized fashion. Such disintegrations are referred to as natural radioactivity. It is also
possible for scientists to smash nuclear particles together and cause nuclear reactions between normally
stable nuclei. These disintegrations are referred to artificial radioactivity. None of the elements above
#92 on the periodic table occur on earth naturally . . . they are all products of artificial radioactivity
(man-made).

Fission and Chain Reactions

Nuclei that are larger than iron-56 become smaller and in the process become more stable. These large
nuclei undergo nuclear reactions in which they break up into two or more smaller nuclei. These reactions
are called fission reactions.

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Conversely, nuclei that are smaller than iron-56 become larger nuclei in order to be more stable. These
nuclei undergo a nuclear reaction in which smaller nuclei join together to form a larger nucleus. Such
nuclear reactions are called fusion reactions. In both fission and fusion, large amounts of energy are given
off in the form of heat, light, and gamma radiation. Nuclear fission was discovered in the late 1930s when
U - 235 nuclides were bombarded with neutrons and were observed to split into two smaller-mass nuclei.

1 _L 235 TT

n + 92 U

141 92 1

56 Ba + 36 Kr + 3 n

The products shown are only one of many sets of products from the disintegration of a U — 235 nucleus.
Over 35 different elements have been observed in the fission products of U — 235.

> fc

n n

When a U — 235 nucleus captures a neutron, it undergoes fission producing two lighter nuclei and three
free neutrons (Figure ??). The production of the free neutrons makes it possible to have a self-sustaining
fission process - a nuclear chain reaction. If at least one of the neutrons goes on to cause another
U - 235 disintegration, the fission will be self-sustaining. If none of the neutrons goes on to cause another
disintegration, the process dies out. If the mass of fissionable material is too small, the neutrons escape
from the mass without causing another reaction and the reaction is said to be subcritical. When the
mass of fissionable material is large enough, at least one of the neutrons will cause another reaction and
the process will continue at a steady rate. This process is said to be critical.

When the critical mass is reached, the neutron will always cause another disintegration.

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Rod x

When the mass is small enough, the neutron
may escape without causing another reaction.
When the mass is large, however, the neutron is
very unlikely to escape.

If enough mass is present for the reaction to escalate rapidly, the heat buildup causes a violent explosion.
This situation is called supercritical. The amount of mass necessary to maintain a chain reaction differs
with each fissionable material and is called that material's critical mass.

A short animation of nuclear fission can be viewed at http://www.classzone.com/cz/books/woc_
07/resources/htmls/ani_chem/chem_f lash/popup. html?layer=act&src=qtiwf_act 129. 1 .xml

Natural and Artificial Radioactivity

There are several ways in which nuclei can undergo reaction and change their identity. Some nuclei are
unstable and spontaneously emit particles and electromagnetic radiation. Such spontaneous disintegration
is known as natural radioactivity.

It is also possible to cause nuclear disintegration by striking a nucleus with another particle. This is
called artificial radioactivity. Ernest Rutherford produced the first induced or artificial transmutation
of elements by bombarding a sample of nitrogen gas with alpha particles.

238 TT 1

239 TT

239_

239„

92 U + n -

^ U -
92

--i e +93 Np -

--I e +94 PU

Irene Joliet-Curie, the daughter of Pierre and Marie Curie, carried out a transmutation that produced
an unstable (radioactive) nucleus from a stable nucleus. She bombarded aluminum with alpha particles,
and after the bombardment was stopped the product continued to emit radiation. It was determined
that the alpha particle bombardment transmuted aluminum-27 nuclei to phosphorus-30 nuclei. The
phosphorus-30 is a positron emitter. Positrons are subatomic particles with the same mass as an electron,
but carry a positive one (+1) charge instead of negative one (-1).

4 TT 27 _, 30„ 1

2 He + 13 A1 -> l5 P + Q n

30 30

15 P "» 14 Sl + 1 P

Fusion

The nuclear reaction shown below is a reaction in which a lithium-7 nucleus combines with a hydrogen-1
nucleus and produces two helium-4 nuclei and a considerable amount of energy.

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17 4

1 H + Li -> 2 2 He + energy

The combined mass of the lithium and hydrogen nuclei is 8.02329 Daltons and the combined mass of the
two helium nuclei is 8.00520 Daltons. The mass lost is 0.01809 Daltons and is accounted for in the energy
that is released. Nuclear reactions, in which two or more lighter-mass nuclei join together to form a single
nucleus, are called fusion reactions or nuclear fusions. Of particular interest are fusion reactions in which
hydrogen nuclei combine to form helium. Hydrogen nuclei are positively charged and repel each other.
The closer the particles come, the greater is the force of repulsion. In order for fusion reactions to occur,
the hydrogen nuclei must have extremely high kinetic energies so the velocities can overcome the forces
of repulsion. These kinetic energies only occur at extreme temperatures such as those that occur in the
cores of the sun and other stars. Nuclear fusion is the power source for the stars where the necessary
temperature to ignite the fusion reaction is provided by massive gravitational pressure.

The energy that comes from the sun and other stars is produced by fusion. (Source: http: //commons.
wikimedia.org/wiki/File: Sun- in-X-ray. Public Domain)

The heat to ignite the fusion reaction in thermonuclear weapons (hydrogen bombs) is provided by a small
fission reaction that is set off inside the mass of hydrogen. It is hoped that the fusion reactions that will
someday occur in fusion reactors will be ignited by multiple lasers.

The conversion of hydrogen to helium in the sun requires several steps, but the net result is the fusion of
four hydrogen-1 nuclei into one helium-4 nucleus.

4 1 H -> 2 He + 2 1 p + energy

In stars more massive than our sun, fusion reactions involving carbon and nitrogen are possible. These
reactions produce more energy than hydrogen fusion reactions.

The exact reactions involved in thermonuclear weapons are secret, but they most likely involve either a
reaction between two hydrogen-2 nuclei or a reaction between a hydrogen-2 nucleus and a hydrogen-3
nucleus.

2 2 4

H + H -» He + energy

2 XT 3 XT 4 XT 1

jH + H -> 2 He + Q n + energy

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Intensive research is now being conducted to develop fusion reactors for electricity generation. The two
major problems slowing up the development is finding a practical means for generating the intense tem-
perature needed and developing a container than won't melt under the conditions of a fusion reaction.
Electricity-producing fusion reactors are still a distant dream.

A short animation of nuclear fusion can be viewed at http://www.classzone.com/cz/books/woc_
07/resources/htmls/ani_chem/chem_f lash/popup. html?layer=act&src=qtiwf _act 130. 1 .xml

Lesson Summary

• In nuclear reactions, the sum of the atomic numbers and the sum of the mass numbers on the two
sides of the equation must be equal.

• Naturally radioactive elements exist in the earth and are either alpha or beta emitters.

• Artificial transmutation of elements can be accomplished by bombarding the nuclei of some elements
with alpha or subatomic particles.

• Nuclear fission refers to the splitting of atomic nuclei.

• Nuclear fusion refers to the joining together to two or more smaller nuclei to form a single nucleus.

Vocabulary

artificial radioactivity Induced radioactivity that is produced by bombarding an element with high-
velocity particles.

chain reaction A multi-stage nuclear reaction that sustains itself in a series of fissions in which the
release of neutrons from the splitting of one atom leads to the splitting of others.

critical mass The smallest mass of a fissionable material that will sustain a nuclear chain reaction at a
constant level.

fission A nuclear reaction in which a heavy nucleus splits into two or more smaller fragments, releasing
large amounts of energy.

fusion A nuclear reaction in which nuclei combine to form more massive nuclei with the simultaneous
release of energy.

natural radioactivity The radioactivity that occurs naturally, as opposed to induced radioactivity. Also
known as spontaneous fission.

Further Reading / Supplementary Links

http : //www. atomicarchive . com/Movies/index_movies . shtml

Review Questions

1. Only one particle is missing from this, equation. What are its atomic and mass numbers?

7 N + 2 He -» .H + ?

2. To what element does the missing particle in question #1 belong?
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3. When a U — 235 nucleus is struck by a neutron, the nucleus may be split into Ce — 144 and Sr — 90
nuclei, also emitting four electrons and two neutrons. Write the equation for this nuclear reaction.

4. Complete the following nuclear equatioqljg supplying the missing particle.

5. Complete the following nuclear equationjjjg supplying the missing particle.

6. Complete the following nuclear equation Jgr supplying the missing particle.

7. Complete the following nuclear equatioj^W supplying ^j^jmissing particle.

80 kg+f ^ 79 AU

8. Complete the following nuclear equation by supgtygng the pissing particle.

? + "» 84 Po + -i e

c j

^Fe

o '

238, .

— — — _u

J 1 ^

1T"~C

K 4 He

CJI

1 H

Mass Number

Use information in the chart above to decide if carbon- 12 nuclei were to be transmuted into other nuclei
that were more stable, would this more likely be accomplished by fission or by fusion?

19.6 Radiation Around Us

Lesson Objectives

The student will:

• compare qualitatively the ionizing and penetration power of a, fi, and y particles.

• calculate the amount of radioactive material that will remain after an integral number of half-lives.

• describe how carbon-14 is used to determine the age of carbon containing objects.

Introduction

All of us are subjected to a certain amount of ionizing radiation every day. This radiation is called
background radiation and comes from a variety of natural and artificial radiation sources. Approximately
82% of background radiation comes from natural sources. These include 1) sources in the earth; naturally
occurring radioactive elements which are incorporated in building materials and also in the human body,
2) sources from space in the form of cosmic rays, and 3) sources in the atmosphere such as radioactive
radon gas released from the earth and radioactive atoms like carbon-14 produced in the atmosphere by
bombardment from high-energy cosmic rays.

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Background Radiation

Approximately 15% of background radiation comes from medical x-rays and nuclear medicine. The re-
maining 3% of background radiation comes from man-made sources such as: smoke detectors, luminous
dials and signs, radioactive contamination due to historical nuclear weapons testing, normal operation of
facilities used for nuclear power and scientific research, emissions from burning fossil fuels (primarily coal-
burning power plants without ash-capture facilities), emissions from the improper disposal of radioactive
materials used in nuclear medicine

Public health agencies do not feel that the level of background radiation is a serious threat to public health,
but they recommend that individuals limit their exposure to ionizing radiation as much as possible . To
this goal, the medical profession has significantly reduced the number of x-rays recommended: skin tests for
tuberculosis are recommended over x-rays and most dentists recommend dental x-rays every other check-up
instead of every check-up.

The Ionizing and Penetration Power of Radiation

With all the radiation from natural and man-made sources, we should quite reasonably be concerned about
how all the radiation might affect our health. The damage to living systems is done by radioactive emissions
when the particles or rays strike tissue, cells, or molecules and alter them. These interactions can alter
molecular structure and function; cells no longer carry out their proper function and molecules, such as
DNA, no longer carry the appropriate information. Large amounts of radiation are very dangerous, even
deadly. The ability of radiation to damage molecules is analyzed in terms of what is called ionizing power.
When a radiation particle interacts with atoms, the interaction can cause the atom to lose electrons and
thus become ionized. The greater the likelihood that damage will occur by an interaction is the ionizing
power of the radiation. Much of the threat from radiation is involved with the ease or difficulty of protecting
oneself from the particles. How thick of a wall do you need to hide behind to be safe? The ability of each
type of radiation to pass through matter is expressed in terms of penetration power. The more material
the radiation can pass through, the greater the penetration power and the more dangerous they are.

Comparing only the three common types of ionizing radiation, alpha particles have the greatest mass.
Alpha particles have approximately four times the mass of a proton or neutron and approximately 8, 000
times the mass of a beta particle. Because of the large mass of the alpha particle, it has the highest ionizing
power and the greatest ability to damage tissue. That same large size of alpha particles, however, makes
them less able to penetrate matter. They collide with molecules very quickly when striking matter, add
two electrons and become a harmless helium atom. Alpha particles have the least penetration power and
can be stopped by a thick sheet of paper. They are also stopped by the outer layer of dead skin on people.
This may seem to remove the threat from alpha particles but only from external sources. In a situation
like a nuclear explosion or some sort of nuclear accident where radioactive emitters are spread around in
the environment, the emitters can be inhaled or taken in with food or water and once the alpha emitter is
inside you, you have no protection at all.

Beta particles are much smaller than alpha particles and therefore, have much less ionizing power (less
ability to damage tissue), but their small size gives them much greater penetration power. Most resources
say that beta particles can be stopped by a one-quarter inch thick sheet of aluminum. Once again, however,
the greatest danger occurs when the beta emitting source gets inside of you.

Gamma rays are not particles but a high energy form of electromagnetic radiation (like x-rays except more
powerful). Gamma rays are energy that has no mass or charge. Gamma rays have tremendous penetration
power and require several inches of dense material (like lead) to shield them. Gamma rays may pass all
the way through a human body without striking anything. They are considered to have the least ionizing
power and the greatest penetration power.

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When researching thicknesses and materials required to stop various types of radiation, different estimates
are encountered. There are apparently two reasons for this: 1) some estimates are based on stopping 95%
of beta particles while other estimates are based on stopping 99% of beta particles, and 2) different beta
particles emitted during nuclear reactions may have very different energies - some very high energy beta
particles may not be stopped by the normal barrier. The safest amount of radiation to avoid damage to
the human body is zero. It isn't possible to be exposed to zero ionizing radiation so the next best goal is
to be exposed to as little as possible.

Definition of Half-Life

During natural radioactive decay, not all atoms of an element are instantaneously changed to atoms of
another element. The decay process takes time and there is value in being able to express the rate at which
a process occurs. In chemical reactions as well as radioactive decay, a useful concept is half-life, which
is the time required for half of the starting material to be consumed. Half-lives can be calculated from
measurements on the change in mass of a nuclide and the time it takes to occur. For a particular group of
radioactive nuclei, it is not possible to know which nuclei will disintegrate or when they will disintegrate.
The only thing we know is that in the time of that substance's half-life, half of the original nuclei will
disintegrate.

Selected Half-Lives

The half-lives of many radioactive isotopes have been determined and they have been found to range from
extremely long half-lives of 10 billion years to extremely short half-lives of fractions of a second (Table
19.2).

Table 19.2: TABLE OF SELECTED HALF-LIVES

ELEMENT

MASS
BER

NUM-

HALF-LIFE

ELEMENT

MASS
BER

NUM-

HALF-LIFE

Uranium

238

4.5 billion
years

Californium

251

800 years

Neptunium

240

1 hour

Nobelium

254

3 seconds

Plutonium

243

5 hours

Carbon

5, 770 years

Americium

246

25 minutes

Carbon

0.74 seconds

The quantity of radioactive nuclei at any given time will decrease to half as much in one half-life. For
example, if there were 100 g of Cf - 251 in a sample at some time, after 800 years, there would be 50 g of
Cf- 251 remaining and after another 800 years, there would only be 25 g remaining.

Radioactive Dating

An ingenious application of half-life studies established a new science of determining ages of materials by
half-life calculations. For geological dating, the decay of U — 238 can be used. The half-life of U — 238
is 4.5 x 10 9 years. The end product of the decay of U — 238 is Pb — 206. After one half-life, a 1.00 gram
sample of uranium will have decayed to 0.50 grams of U — 238 and 0.43 grams of Pb - 206. By comparing
the amount of U - 238 to the amount of Pb - 206 in a sample of uranium mineral, the age of the mineral
can be estimated. Present day estimates for the age of the Earth's crust from this method is at least 4
billion years.

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Organic material is radioactively dated using the long-lived nuclide of carbon, carbon- 14. This method
of determining the age of organic material was given the name radiocarbon dating. The carbon dioxide
consumed by living systems contains a certain concentration of 14 C02- When the organism dies, the
acquisition of carbon-14 stops but the decay of the C — 14 in the body continues. As time goes by, the
ratio of C — 14 to C — 12 decreases at a rate determined by the half-life of C — 14. Using half-life equations,
the time since the organism died can be calculated. These procedures have been used to determine the age
of organic artifacts and determine, for instance, whether art works are real or fake.

Lesson Summary

• 82% of background radiation comes from natural sources. These include 1) sources in the earth;
naturally occurring radioactive elements which are incorporated in building materials and also in the
human body, 2) sources from space in the form of cosmic rays, and 3) sources in the atmosphere such
as radioactive radon gas released from the earth and radioactive atoms like carbon-14 produced in
the atmosphere by bombardment from high-energy cosmic rays.

• Approximately 15% of background radiation comes from medical x-rays and nuclear medicine. The
remaining 3% of background radiation comes from man-made sources such as: smoke detectors,
luminous dials and signs, radioactive contamination due to historical nuclear weapons testing, normal
operation of facilities used for nuclear power and scientific research, emissions from burning fossil fuels
(primarily coal-burning power plants without ash-capture facilities), emissions from the improper
disposal of radioactive materials used in nuclear medicine.

• Of the three common nuclear emissions, alpha particles produce the greatest damage to cells and
molecules but are the least penetrating. Gamma rays are the most penetrating but generated the
least damage.

• C — 14 dating procedures have been used to determine the age of organic artifacts.

Vocabulary

background radiation Radiation that comes from environment sources including the earth's crust, the
atmosphere, cosmic rays, and radioisotopes. These natural sources of radiation account for the largest
amount of radiation received by most people.

half-life The half-life of a radioactive substance is the time interval required for a quantity of material
to decay to half its original value.

Further Reading / Supplementary Links
Review Questions

1 . Which of the three common emissions from radioactive sources requires the heaviest shielding?

2. The half-life of radium-226 is about 1600 years. How many grams of a 2.00 gram sample will remain
after 4800 years?

3. Sodium-24 has a half-life of about 15 hours. How much of an 16.0 grams sample of sodium-24 will
remain after 60.0 hours?

4. A radioactive isotope decayed from 24.0 grams to 0.75 grams in 40.0 years. What is the half-life of
the isotope?

5. What nuclide is commonly used in the dating of organic artifacts?

6. Why does an ancient wood artifact contain less carbon-14 than a piece of lumber sold today?

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7. The half-life of C — 14 is about 5, 700 years. An organic relic is found to contain C — 14 and C — 12
in a ratio that is about one-eighth as great as the ratio in the atmosphere. What is the approximate
age of the relic?

8. Even though gamma rays are much more penetrating than alpha particles, it is the alpha particles
that are more likely to cause damage to an organism. Explain why this is true.

9. The radioactive isotope calcium-47 has been used in the study of bone metabolism; radioactive
iron-59 has been used in the study of red blood cell function; iodine-131 has been used in both
diagnosis and treatment of thyroid problems. Suggest a reason why these particular elements were
chosen for use with the particular body function.

19.7 Applications of Nuclear Energy

Lesson Objectives

The student will:

• trace the energy transfers that occur in a nuclear power reactor from the binding energy of the nuclei
to the electricity that leaves the plant.

• define the term "breeder reactor."

• list some medical uses of nuclear energy.

Introduction

It is unfortunate that when the topics of radioactivity and nuclear energy come up, most thoughts probably
go to weapons of war. The second thought might be about the possibility of nuclear energy contributing to
the solution of the energy crisis. Nuclear energy, however, has many applications beyond bombs and the
generation of electricity. Radioactivity has huge applications in scientific research, several fields of medicine
both in terms of imaging and in terms of treatment, industrial processes, some very useful appliances, and
even in agriculture.

Fission Reactors

A nuclear reactor is a device in which a nuclear chain reaction is carried out at a controlled rate. When
the controlled chain reaction is a fission reaction, the reactor is called a fission reactor. Fission reactors are
used primarily for the production of electricity although there are a few fission reactors used for military
purposes and for research. The great majority of electrical generating systems all follow a reasonably
simple design. The electricity is produced by spinning a coil of wire inside a magnetic field.

When the loop is spun, electric current is produced. The direction of the electric current is in one end of
the loop and out the other end. The machine built to accomplish this task is called an electric generator

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(see figure above). Another machine usually involved in the production of electric current is a turbine.
Although actual turbines can get very complicated, the basic idea is simple. Most of you have seen a child's
pinwheel toy. When you blow through the fan blades of the pin wheel, it spins. A turbine is a pipe with
many fan blades attached to an axle that runs through the pipe.

When a fluid (air, steam, water) is forced through the pipe, it spins the fan blades which in turn spin the
axle. To generate electricity, the axle of a turbine is attached to the loop of wire in a generator. When a
fluid is forced through the turbine, the fan blades turn, the turbine axle turns, and the loop of wire inside
the generator turns, thus generating electricity.

A steam turbine. (Source: CK-12 Foundation. CC-BY-SA)

The essential difference in various kinds of electrical generating systems is the method used to spin the
turbine. For a wind generator, the turbine is a windmill. In a geothermal generator, steam from a geyser
is forced through the turbine. In hydroelectric generating plants, water falling over a dam passes through
the turbine and spins it. In fossil fuel (coal, oil, natural gas) generating plants, the fossil fuel is burned
and the heat is used to boil water into steam and then the steam passes through the turbine and makes it
spin. In a fission reactor generating plant, a fission reaction is used to boil the water into steam and the
steam passes through the turbine to make it spin. Once the steam is generated by the fission reaction, a
nuclear power plant is essentially the same as a fossil fuel plant.

Naturally occurring uranium is composed almost totally of two uranium isotopes. It contains more than
99% uranium-238 and less than 1% uranium-235. It is the uranium-235, however, that is fissionable
(will undergo fission). In order for uranium to be used as fuel in a fission reactor, the percentage of
uranium-235 must be increased, usually to about 3%. (Uranium in which the U — 235 content is more
than 1% is called enriched uranium.) Somehow, the two isotopes must be separated so that enriched
uranium is available for use as fuel. Separating the isotope by chemical means (chemical reactions) is
not successful because the isotopes have exactly the same chemistry. Remember that chemical reactions
are controlled by the electron configuration of the atom and all the isotopes of an element have the same
electron configuration and hence the same chemistry. The only essential difference between U - 238 and
U - 235 are their atomic masses and as a result the separation of the two isotopes will require a physical
means that takes advantage of their difference in mass. The modern physical means used to separate the
isotopes of uranium involve a gas centrifuge. Research separation of isotopes can also be accomplished
with a mass spectrograph. You may wish to research these techniques on your own.

Once the supply of U — 235 is acquired, it is placed in a series of long cylindrical tubes called fuel rods.
These fuel cylinders are bundled together with control rods (Figure ??) made of neutron-absorbing
material. The amount of U— 235 in all the fuel rods taken together is adequate to carry on a chain reaction
but is less than the critical mass. (In the United States, all public nuclear power plants contain less than
a critical mass of U — 235 and therefore, could never produce a nuclear explosion. The amount of heat

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generated by the chain reaction is controlled by the rate at which the nuclear reaction occurs. The rate of
the nuclear reaction is dependent on how many neutrons are emitted by one U — 235 nuclear disintegration
and strike a new U — 235 nucleus to cause another disintegration. The purpose of the control rods is to
absorb some of the neutrons and thus stop them from causing further disintegrations. The control rods
can be raised or lowered into the fuel rod bundle. When the control rods are lowered all the way into the
fuel rod bundle, they absorb so many neutrons that the chain reaction essentially stops. When more heat
is desired, the control rods are raised so they catch fewer neutrons, the chain reaction speeds up and more
heat is generated. The control rods are operated in a fail-safe system so that power is necessary to hold
them up; and during a power failure, gravity will pull the control rods down into shut off position.

Hot coolant

Control rods
made of neutron
absorbing
material

Uranium
fuel rods

Incoming coolant (cooler)

U - 235 nuclei can capture neutrons and disintegrate more efficiently if the neutrons are moving slower
than the speed at which they are released. Fission reactors use a moderator surrounding the fuel rods to
slow down the neutrons. Water is not only a good coolant but also a good moderator so a common type of
fission reactor has the fuel core submerged in a huge pool of water. This type of reaction is called a Light
Water Reactor or LWR. All public electricity generating fission reactors in the United States are LWRs.

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Reinforced concrete
containment structure

Steel contarment
shell

PRESSURIZED

WATER REACTOR

(PWR)

Steam spins
the turbine

Note how everything
is self contained.
Nothing leaves the
system except
heat energy.

Steam is cooled
back to water

You can follow the operation of an electricity-generating fission reactor in Figure ??. The reactor core is
submerged in a pool of water. The heat from the fission reaction heats the water and the water is pumped
into a heat exchanger container where the heated water boils the water in the heat exchanger. The steam
from there is forced through a turbine which spins a generator and produces electricity. After the water
passes through the turbine, it is condensed back to liquid water and pumped back to the heat exchanger.

In the United States, heavy opposition to the use of nuclear energy was mounted in the late 1960s and
early 1970s. Every environmentalist organization in the US opposed the use of nuclear energy and the
constant pressure from environmentalist groups brought increased public fear and therefore, opposition.
This is not true today; at least one environmental leader has published a paper in favor of nuclear powered
electricity generation.

In 1979, a reactor core meltdown at Pennsylvania's Three Mile Island nuclear power plant reminded the
entire country of the dangers of nuclear radiation. The concrete containment structure (six feet thick walls
of reinforced concrete), however, did what it was designed to do - prevent radiation from escaping into
the environment. Although the reactor was shut down for years, there were no injuries or deaths among
nuclear workers or nearby residents . Three Mile Island was the only serious accident in the entire history
of civilian power plants (103 plants operating for 40 years) in the United States. There has never been a
single injury or death due to radiation in any public nuclear power plant in the U.S. The accident at Three
Mile Island did, however, frighten the public so that there has not been a nuclear plant built in the U.S.
since the accident.

The 103 nuclear power plants operating in the U.S. deliver approximately 19.4% of American electricity
with zero greenhouse gas emission. There are 600 coal-burning electric plants in the US delivering 48.5%
of American electricity and producing 2 billion tons of CO2 annually, accounting for 40% of U.S. CO2
emissions and 10% of global emissions. These coal burning plants also produce 64% of the sulfur dioxide
emissions, 26% of the nitrous oxide emissions, and 33% of mercury emissions. 4

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Breeder Reactors

U - 235 is the only naturally occurring fissile isotope and it constitutes less than 1% of naturally occurring
uranium. It has been projected that the world's supply of U — 235 will be exhausted in less than 200 years.
It is possible, however, to convert U — 238 to a fissionable isotope which will function as a fuel for nuclear
reactors. The fissionable isotope is plutonium-239 and is produced by the following series of reactions.

238 XT 1
92 U + n

239

, 239_ T

239

-I e+ 94 PU

The final product from this series of reactions is plutonium-239 which has a half-life of 24, 000 years and
is another nuclear reactor fuel. This series of reactions can be made to occur inside an operating nuclear
reactor by replacing some of the control rods with rods of U - 238. As the nuclear decay process proceeds
inside the reactor, it produces more fuel than it uses. It would take about 20 such breeder reactors to
produce enough fuel to operate one addition reactor. The use of breeder reactors would extend the fuel
supply a hundred fold. The problem with breeder reactors, however, is that plutonium is an extremely
deadly poison and unlike ordinary fission reactors, it is possible for out-of-control breeder reactors to
explode. None of the civilian nuclear power plants in the U. S. are breeder reactors.

Radiation Detectors

A variety of methods have been developed to detect nuclear radiation. One of the most commonly used
instruments for detecting radiation and measuring the rate is the Geiger counter.

A Geiger Counter. (Source: http://commons.wikipedia.Org/wiki/File:Geiger.png. GNU Free Doc-
umentation)

Cathode

Ionizing Radiation

The detecting component of the Geiger counter is the Geiger-Muller tube. This tube is a cylinder filled
with an inert gas and it has a window in one end made of porous material that will not allow the inert gas
to escape but will allow radiation particles to enter. A conducting wire extends into the center of the tube

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and is electrically insulated from the tube where it passes through the wall. The wire and walls are part of
an electric circuit with a potential difference between the walls and the wire. Electric current does not flow
because the circuit is not complete. The inert gas does not conduct electricity and therefore a gap exists
in the circuit. When a radiation particle enters the tube through the window, the particle creates a line of
ionized gas particles along its path through the tube. The line of ions does conduct electric current and
an electric current will flow along the ionized path. The ions only exist for a very short period because the
ions of inert gas will quickly regain the lost electrons and become atoms again - which do not conduct. The
result is a very short burst of electric current whenever a radiation particle passes through the tube. The
control box provides the electric potential for the tube and also provides some means for demonstrating
the burst of current. Some machines simply make a clicking sound for each burst of current while others
may provide a dial or a digital meter.

Other methods used for the detection of nuclear radiation include, 1) scintillation counters - a screen
coated with a material that gives off a small flash of light when struck by a particle, 2) cloud chambers
(see image below) - a chamber of supersaturated gas that produce a condensation trail along the path of
a radiation particle, and 3) bubble chambers - a chamber of superheated liquid that produces a trail of
bubbles along the path of a radiation particle.

Cloud Chamber showing vapor trails produced by sub-atomic particles. (Source: http://www.nasa.gov/
multimedia/imagegallery/image_f eature_928_prt .htm. Public Domain)

Cloud chambers and bubble chambers have an additional value because the vapor trails or bubble trails left
by the nuclear radiation particle are long-lasting enough to be photographed and therefore can be studied
in great detail.

Particle Accelerators

In the early 1900's, the use of alpha particles for bombarding low atomic number elements became a
common practice. Researchers found that the alpha particles were absorbed by the nuclei and a proton
was ejected. This was the first artificially caused transmutation of one element into another. In order to
continue these bombardments with alpha particles or protons, the speed of the bombarding particle had
to be increased. Several machines were devised to accelerate the particles to the required speeds.

The cyclotron was developed by Ernest Lawrence in 1930 and used to accelerate charged particles so they
would have sufficient energy to enter the nuclei of target atoms. A cyclotron consists of two hollow half

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cylinders called "dees" because of their D-shapes. (Like a huge birthday cake which has been cut in half
and the two halves separated by a short distance.)

Cyclotron.

The two "dees" have opposite charges with a potential difference of at least 50, 000 volts and the charges
on the dees can be rapidly reversed so that each dee alternately becomes positive and then negative.
The cyclotron also has a powerful magnetic field passing through it so that moving charged particles will
be caused to travel in a curved path. Charged particles produced from a source in the center of the area
between the dees are attracted first into one dee and then into the other as the charge on the dees alternate.
As the particle moves back and forth in the dees, it is caused to follow a curved path due to the magnetic
field. The motion of the particle is that of a spiral with ever increasing speed. As the circular path of the
particle nears the outside edge of the cyclotron, it is allowed to exit through a window and strikes whatever
target is placed outside the window.

A linear accelerator is a long series of tubes which are connected to a source of high frequency alternating
voltage. As the charged particles leave each tube, the charge on the tubes are altered so the particle is
repelled from the tube it is leaving and attracted to the tube it is approaching. In this way, the particle
is accelerated between every pair of tubes. The largest linear accelerators are located at the Fermilab in
Illinois and at Stanford University.

A number of the elements listed in the periodic table are not found in nature. These elements may never
have been present on earth or since they have short half- lives, they may have originally been present but
have completely decayed to more stable elements. These elements include all elements with atomic numbers
greater than 92 plus technetium (#43) and promethium (#61). The transuranium elements (those with
atomic numbers greater than 92) are all made man elements and many of them were produced in the
cyclotron in the radiation laboratory at the University of California at Berkeley under the direction of
Glenn Seaborg (Figure ??). Some very rare elements presence in the earth are assumed to be due not
from the original material present in the earth but rather as daughter products of other disintegrating
nuclei.

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The following web site contains a history of the discovery (creation) of the transuranium
elements and includes some of the reactions used to produce them.

The Search for "Heavy" Elements (http://www.lbl.g0v/abc/wallchart/chapters/O8/O.html)

Nuclear Medicine

The held of nuclear medicine has expanded greatly in the last twenty years. A great deal of the expansion
has come in the area of imaging. This section will focus on nuclear medicine involving the types of nuclear
radiation introduced in this chapter. The x-ray imaging systems will not be covered.

Radioiodine (1-131) Therapy involves imaging and treatment of the thyroid gland. The thyroid gland
is a gland in the neck that produces two hormones that regulate metabolism. In some individuals, this
gland becomes overactive and produces too much of these hormones. The treatment for this problem uses
radioactive iodine (1-131) which is produced for this purpose in research hssion reactors or by neutron
bombardment of other nuclei.

The thyroid gland uses iodine in the process of its normal function. Any iodine in food that enters the
bloodstream is usually removed by, and concentrated in the thyroid gland. When a patient suffering from
an overactive thyroid swallows a small pill containing radioactive iodine, the 1-131 is absorbed into the
bloodstream just like non-radioactive iodine and follows the same process to be concentrated in the thyroid.
The concentrated emissions of nuclear radiation in the thyroid destroy some of the gland's cells and control
the problem of the overactive thyroid.

Smaller doses of 1-131 (too small to kill cells) are also used for purposes of imaging the thyroid. Once the
iodine is concentrated in the thyroid, the patient lays down on a sheet of him and the radiation from the
1-131 makes a picture of the thyroid on the him. The half- life of iodine- 131 is approximately 8 days so after
a few weeks, virtually all of the radioactive iodine is out of the patient's system. During that time, they
are advised that they will set off radiation detectors in airports and will need to get special permission to
fly on commercial flights.

Positron Emission tomography or PET scan is a type of nuclear medicine imaging. Depending on the
area of the body being imaged, a radioactive isotope is either injected into a vein, swallowed by mouth,
or inhaled as a gas. When the radioisotope is collected in the appropriate area of the body, the gamma
ray emissions are detected by a PET scanner (often called a gamma camera) which works together with
a computer to generate special pictures providing details on both the structure and function of various
organs. PET scans are used to:

• detect cancer

• determine the amount of cancer spread

• assess the effectiveness of treatment plans

• determine blood flow to the heart muscle

• determine the effects of a heart attack

• evaluate brain abnormalities such as tumors and memory disorders

• map brain and heart function

External Beam Therapy (EBT) is a method of delivering a high energy beam of radiation to the precise
location of a patient's tumor. These beams can destroy cancer cells and with careful planning, NOT kill
surrounding cells. The concept is to have several beams of radiation, each of which is sub-lethal, enter
the body from different directions. The only place in the body where the beam would be lethal is at the
point where all the beams intersect. Before the EBT process, the patient is three-dimensionally mapped
using CT scans and x-rays. The patient receives small tattoos to allow the therapist to line up the beams

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exactly. Alignment lasers are used to precisely locate the target. The radiation beam is usually generated
with a linear accelerator. EBT is used to treat the following diseases as well as others:

• breast cancer

• colorectal cancer

• head and neck cancer

• lung cancer

• prostate cancer

Nuclear Weapons

Nuclear weapons are of two basic types; fission bombs using supercritical masses of either U-235 or Pu-239
and fusion bombs using heavy isotopes of hydrogen. The fission bombs were called atomic bombs (a
misnomer since the energy comes from the nucleus) and fusion bombs are called thermonuclear bombs.
Fission bombs use two or more subcritical masses of fissile material separated by enough distance that
they don't become critical, and surrounded by conventional explosives. The conventional explosives are
detonated to drive the subcritical masses of fissile material toward the center of the bomb and when these
masses are slammed together, they form a supercritical mass and a nuclear explosion ensues. Hydrogen
bombs (fusion) are detonated by using a small fission explosion to compress and heat a mass of deuterium
or deuterium and tritium to the point that a fusion reaction ignites.

Castle-Romeo nuclear explosion. (Source: http://en.wikipedia.Org/wiki/File:Castle_romeo2.jpg.
Public Domain)

Nuclear weapons are power-rated by comparison to the weight of conventional explosives (TNT) that would
produce an equivalent explosion. For example, a nuclear device that produces an explosion equivalent to
1,000 tons (2,000,000 pounds) of TNT would be called a 1-kiloton bomb. The atomic bomb detonated
at Hiroshima near the end of WWII was a 13-kiloton bomb that used 130 pounds of U — 235. It has been
estimated that this weapon was very inefficient and that less than 1.5% of the fissile material actually
fissioned. The atomic bomb detonated at Nagasaki was a 21-kiloton weapon that used 14 pounds of
Pu — 239. The two bombs set off in Japan would be considered very small bombs by later standards.
Bombs that were tested later were measured not by kilotons but by megatons (million tons) of TNT. The
largest bomb ever set off was a 50-megaton fusion weapon tested by the Soviet Union in the 1950s.

The extensive death and destruction caused by these weapons comes from four sources. The tremendous
heat released by the explosion heats the air so hot and so fast that the air expansion creates a wind in
excess of 200 miles/hour - many times stronger than the strongest hurricane. The blast force from this
wind completely destroys all but the strongest buildings for several miles from ground zero. The second
source of damage is from fires ignited by the heat from the fireball at the center of the explosion. The
fireball in the 50-megaton test was estimated to be four miles in diameter. The third source of injury
is the intense nuclear radiation (primarily gamma rays) which are instantly lethal to exposed people for

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several miles. The final source of injury and possibly death conies from the radioactive fall-out which may
be several tons of radioactive debris and may fall up to 300 miles or more away. This fall-out can cause
sickness and death for many years.

You should note that the distances given in the above material are guesses and estimates. The only actual
data known are from the two (now considered small) bombs detonated in Japan and some above ground
tests. The actual death and destruction from a 50-megaton bomb detonated in a heavily populated area
is not known.

Lesson Summary

• The fission of U — 235 or Pu — 239 is used in nuclear reactors.

• The critical mass is the amount of fissile material that will maintain a chain reaction.

• Nuclear radiation also has many medical uses.

Vocabulary

control rods Control rods are made of chemical elements capable of absorbing many neutrons and are
used to control the rate of a fission chain reaction in a nuclear reactor.

cyclotron A cyclotron is a type of particle accelerator.

fall out Fall out is radioactive dust hazard from a nuclear explosion, so named because it "falls out" of
the atmosphere where it was spread by the explosion.

fissile A fissile substance is a substance capable of sustaining a chain reaction of nuclear fission.

fissionable A fissionable material is material capable of undergoing fission.

Geiger counter A Geiger counter is an instrument used to detect radiation, usually alpha and beta
radiation, but some models can also detect gamma radiation.

isotope Nuclei with the same number of protons but different numbers of neutrons.

linear accelerator A linear accelerator is a linear electrical device for the acceleration of subatomic
particles.

moderator A neutron moderator is a medium which reduces the velocity of fast neutrons; commonly
used moderators are regular (light) water, solid graphite, and heavy water.

nuclear pile A nuclear pile is a nuclear reactor.

Further Reading / Supplementary Links

• Nuclear Power VS. Other Sources of Power, Neil M. Cabreza, Department of Nuclear Engineering,
University of California, Berkeley, NE-161 Report. Available at http://www.nuc.berkeley.edu/
thyd/nel61/ncabreza/sources .

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www.hps.org/publicinformation/ate/catlO.htnil

www.sciencemag.org/cgi/content/full/309/5732/233

www.hrd.qut.edu.au/toolkit/Faqs/radiation.jsp

www.radiationnetwork.com/RadiationNetwork.htm

http : //www . iaea . org/NewsCenter/Features/Chernobyl- 15

http : //www . don . wa . gove/ehp/rp/f actsheets-pd/f slO

http : //www . radscihealth . org/rsh/About

http : //nrc . gov/reading-rm/doc-collections

http : //www . world-nuclear . org/inf o/Chernobyl

http : //www . iaea . org/NewsCenter/Features/Chernobyl- 15

http : //www . nuclearweaponarchive . org/Russia/Tsarbomba . html

http : //www . atomicarchive . com/effects/index . shtml

Review Questions

1. What is the primary physical difference between a nuclear electricity generating plant and a coal-
burning electricity generating plant?

2. What do the control rods in a nuclear reactor do and how do they do it?

3. What is a breeder reactor?

4. Name two types of particle accelerators.

5. In the medical use of radioactivity, what does EBT stand for?

6. Is it possible for a nuclear explosion to occur in a nuclear reactor? Why or why not?

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